YES We show the termination of the TRS R: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) a____(X,nil()) -> mark(X) a____(nil(),X) -> mark(X) a__and(tt(),X) -> mark(X) a__isList(V) -> a__isNeList(V) a__isList(nil()) -> tt() a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) a__isNeList(V) -> a__isQid(V) a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) a__isNePal(V) -> a__isQid(V) a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) a__isPal(V) -> a__isNePal(V) a__isPal(nil()) -> tt() a__isQid(a()) -> tt() a__isQid(e()) -> tt() a__isQid(i()) -> tt() a__isQid(o()) -> tt() a__isQid(u()) -> tt() mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) mark(and(X1,X2)) -> a__and(mark(X1),X2) mark(isList(X)) -> a__isList(X) mark(isNeList(X)) -> a__isNeList(X) mark(isQid(X)) -> a__isQid(X) mark(isNePal(X)) -> a__isNePal(X) mark(isPal(X)) -> a__isPal(X) mark(nil()) -> nil() mark(tt()) -> tt() mark(a()) -> a() mark(e()) -> e() mark(i()) -> i() mark(o()) -> o() mark(u()) -> u() a____(X1,X2) -> __(X1,X2) a__and(X1,X2) -> and(X1,X2) a__isList(X) -> isList(X) a__isNeList(X) -> isNeList(X) a__isQid(X) -> isQid(X) a__isNePal(X) -> isNePal(X) a__isPal(X) -> isPal(X) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(__(X,Y),Z) -> mark#(X) p3: a____#(__(X,Y),Z) -> a____#(mark(Y),mark(Z)) p4: a____#(__(X,Y),Z) -> mark#(Y) p5: a____#(__(X,Y),Z) -> mark#(Z) p6: a____#(X,nil()) -> mark#(X) p7: a____#(nil(),X) -> mark#(X) p8: a__and#(tt(),X) -> mark#(X) p9: a__isList#(V) -> a__isNeList#(V) p10: a__isList#(__(V1,V2)) -> a__and#(a__isList(V1),isList(V2)) p11: a__isList#(__(V1,V2)) -> a__isList#(V1) p12: a__isNeList#(V) -> a__isQid#(V) p13: a__isNeList#(__(V1,V2)) -> a__and#(a__isList(V1),isNeList(V2)) p14: a__isNeList#(__(V1,V2)) -> a__isList#(V1) p15: a__isNeList#(__(V1,V2)) -> a__and#(a__isNeList(V1),isList(V2)) p16: a__isNeList#(__(V1,V2)) -> a__isNeList#(V1) p17: a__isNePal#(V) -> a__isQid#(V) p18: a__isNePal#(__(I,__(P,I))) -> a__and#(a__isQid(I),isPal(P)) p19: a__isNePal#(__(I,__(P,I))) -> a__isQid#(I) p20: a__isPal#(V) -> a__isNePal#(V) p21: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p22: mark#(__(X1,X2)) -> mark#(X1) p23: mark#(__(X1,X2)) -> mark#(X2) p24: mark#(and(X1,X2)) -> a__and#(mark(X1),X2) p25: mark#(and(X1,X2)) -> mark#(X1) p26: mark#(isList(X)) -> a__isList#(X) p27: mark#(isNeList(X)) -> a__isNeList#(X) p28: mark#(isQid(X)) -> a__isQid#(X) p29: mark#(isNePal(X)) -> a__isNePal#(X) p30: mark#(isPal(X)) -> a__isPal#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p13, p14, p15, p16, p18, p20, p21, p22, p23, p24, p25, p26, p27, p29, p30} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(nil(),X) -> mark#(X) p3: mark#(isPal(X)) -> a__isPal#(X) p4: a__isPal#(V) -> a__isNePal#(V) p5: a__isNePal#(__(I,__(P,I))) -> a__and#(a__isQid(I),isPal(P)) p6: a__and#(tt(),X) -> mark#(X) p7: mark#(isNePal(X)) -> a__isNePal#(X) p8: mark#(isNeList(X)) -> a__isNeList#(X) p9: a__isNeList#(__(V1,V2)) -> a__isNeList#(V1) p10: a__isNeList#(__(V1,V2)) -> a__and#(a__isNeList(V1),isList(V2)) p11: a__isNeList#(__(V1,V2)) -> a__isList#(V1) p12: a__isList#(__(V1,V2)) -> a__isList#(V1) p13: a__isList#(__(V1,V2)) -> a__and#(a__isList(V1),isList(V2)) p14: a__isList#(V) -> a__isNeList#(V) p15: a__isNeList#(__(V1,V2)) -> a__and#(a__isList(V1),isNeList(V2)) p16: mark#(isList(X)) -> a__isList#(X) p17: mark#(and(X1,X2)) -> mark#(X1) p18: mark#(and(X1,X2)) -> a__and#(mark(X1),X2) p19: mark#(__(X1,X2)) -> mark#(X2) p20: mark#(__(X1,X2)) -> mark#(X1) p21: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p22: a____#(X,nil()) -> mark#(X) p23: a____#(__(X,Y),Z) -> mark#(Z) p24: a____#(__(X,Y),Z) -> mark#(Y) p25: a____#(__(X,Y),Z) -> a____#(mark(Y),mark(Z)) p26: a____#(__(X,Y),Z) -> mark#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33, r34, r35, r36, r37, r38, r39, r40 Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a____#_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (1,27) ___A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (22,2) mark_A(x1) = ((1,0),(1,1)) x1 + (0,48) a_____A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (22,3) nil_A() = (5,22) mark#_A(x1) = ((1,0),(1,1)) x1 + (4,50) isPal_A(x1) = x1 + (35,49) a__isPal#_A(x1) = ((1,0),(1,1)) x1 + (2,133) a__isNePal#_A(x1) = ((1,0),(1,1)) x1 + (1,2) a__and#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (1,2) a__isQid_A(x1) = ((1,0),(1,1)) x1 + (7,49) tt_A() = (6,49) isNePal_A(x1) = x1 + (8,1) isNeList_A(x1) = ((1,0),(0,0)) x1 + (8,2) a__isNeList#_A(x1) = ((1,0),(1,1)) x1 + (1,27) a__isNeList_A(x1) = ((1,0),(1,0)) x1 + (8,50) isList_A(x1) = ((1,0),(0,0)) x1 + (10,1) a__isList#_A(x1) = ((1,0),(1,1)) x1 + (9,60) a__isList_A(x1) = ((1,0),(1,0)) x1 + (10,52) and_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (3,1) a__and_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (3,2) a__isNePal_A(x1) = x1 + (8,51) a__isPal_A(x1) = ((1,0),(1,1)) x1 + (35,50) a_A() = (6,50) e_A() = (0,50) i_A() = (0,50) o_A() = (7,1) u_A() = (6,0) isQid_A(x1) = x1 + (7,1) precedence: __ > mark = a____ = nil = isPal = a__isPal = i > a > a__isNeList# > isNePal = a__isList# = a__and = a__isNePal > isNeList = a__isNeList = isList = a__isList > a____# > mark# = a__isPal# = a__isNePal# > tt = e = u > a__isQid = isQid > a__and# > o > and partial status: pi(a____#) = [] pi(__) = [] pi(mark) = [] pi(a____) = [] pi(nil) = [] pi(mark#) = [1] pi(isPal) = [] pi(a__isPal#) = [1] pi(a__isNePal#) = [1] pi(a__and#) = [] pi(a__isQid) = [1] pi(tt) = [] pi(isNePal) = [] pi(isNeList) = [] pi(a__isNeList#) = [1] pi(a__isNeList) = [] pi(isList) = [] pi(a__isList#) = [1] pi(a__isList) = [] pi(and) = [] pi(a__and) = [] pi(a__isNePal) = [] pi(a__isPal) = [] pi(a) = [] pi(e) = [] pi(i) = [] pi(o) = [] pi(u) = [] pi(isQid) = [1] The next rules are strictly ordered: p10 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(nil(),X) -> mark#(X) p3: mark#(isPal(X)) -> a__isPal#(X) p4: a__isPal#(V) -> a__isNePal#(V) p5: a__isNePal#(__(I,__(P,I))) -> a__and#(a__isQid(I),isPal(P)) p6: a__and#(tt(),X) -> mark#(X) p7: mark#(isNePal(X)) -> a__isNePal#(X) p8: mark#(isNeList(X)) -> a__isNeList#(X) p9: a__isNeList#(__(V1,V2)) -> a__isNeList#(V1) p10: a__isNeList#(__(V1,V2)) -> a__isList#(V1) p11: a__isList#(__(V1,V2)) -> a__isList#(V1) p12: a__isList#(__(V1,V2)) -> a__and#(a__isList(V1),isList(V2)) p13: a__isList#(V) -> a__isNeList#(V) p14: a__isNeList#(__(V1,V2)) -> a__and#(a__isList(V1),isNeList(V2)) p15: mark#(isList(X)) -> a__isList#(X) p16: mark#(and(X1,X2)) -> mark#(X1) p17: mark#(and(X1,X2)) -> a__and#(mark(X1),X2) p18: mark#(__(X1,X2)) -> mark#(X2) p19: mark#(__(X1,X2)) -> mark#(X1) p20: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p21: a____#(X,nil()) -> mark#(X) p22: a____#(__(X,Y),Z) -> mark#(Z) p23: a____#(__(X,Y),Z) -> mark#(Y) p24: a____#(__(X,Y),Z) -> a____#(mark(Y),mark(Z)) p25: a____#(__(X,Y),Z) -> mark#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19, p20, p21, p22, p23, p24, p25} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(__(X,Y),Z) -> mark#(X) p3: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p4: a____#(__(X,Y),Z) -> a____#(mark(Y),mark(Z)) p5: a____#(__(X,Y),Z) -> mark#(Y) p6: mark#(__(X1,X2)) -> mark#(X1) p7: mark#(__(X1,X2)) -> mark#(X2) p8: mark#(and(X1,X2)) -> a__and#(mark(X1),X2) p9: a__and#(tt(),X) -> mark#(X) p10: mark#(and(X1,X2)) -> mark#(X1) p11: mark#(isList(X)) -> a__isList#(X) p12: a__isList#(V) -> a__isNeList#(V) p13: a__isNeList#(__(V1,V2)) -> a__and#(a__isList(V1),isNeList(V2)) p14: a__isNeList#(__(V1,V2)) -> a__isList#(V1) p15: a__isList#(__(V1,V2)) -> a__and#(a__isList(V1),isList(V2)) p16: a__isList#(__(V1,V2)) -> a__isList#(V1) p17: a__isNeList#(__(V1,V2)) -> a__isNeList#(V1) p18: mark#(isNeList(X)) -> a__isNeList#(X) p19: mark#(isNePal(X)) -> a__isNePal#(X) p20: a__isNePal#(__(I,__(P,I))) -> a__and#(a__isQid(I),isPal(P)) p21: mark#(isPal(X)) -> a__isPal#(X) p22: a__isPal#(V) -> a__isNePal#(V) p23: a____#(__(X,Y),Z) -> mark#(Z) p24: a____#(X,nil()) -> mark#(X) p25: a____#(nil(),X) -> mark#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33, r34, r35, r36, r37, r38, r39, r40 Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a____#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,10) ___A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (3,6) mark_A(x1) = ((1,0),(1,1)) x1 + (0,15) a_____A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (3,9) mark#_A(x1) = ((1,0),(1,1)) x1 + (4,5) and_A(x1,x2) = x1 + ((1,0),(0,0)) x2 a__and#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (4,5) tt_A() = (5,6) isList_A(x1) = ((1,0),(1,1)) x1 a__isList#_A(x1) = ((1,0),(0,0)) x1 + (1,10) a__isNeList#_A(x1) = ((1,0),(0,0)) x1 + (1,9) a__isList_A(x1) = ((1,0),(1,1)) x1 + (0,8) isNeList_A(x1) = ((1,0),(1,1)) x1 + (0,5) isNePal_A(x1) = x1 a__isNePal#_A(x1) = ((1,0),(0,0)) x1 a__isQid_A(x1) = x1 isPal_A(x1) = ((1,0),(1,1)) x1 + (0,2) a__isPal#_A(x1) = x1 + (0,1) nil_A() = (6,4) a__and_A(x1,x2) = x1 + ((1,0),(0,0)) x2 a__isNeList_A(x1) = ((1,0),(1,1)) x1 + (0,7) a__isNePal_A(x1) = x1 + (0,1) a__isPal_A(x1) = ((1,0),(1,1)) x1 + (0,16) a_A() = (6,7) e_A() = (6,7) i_A() = (6,7) o_A() = (6,7) u_A() = (6,7) isQid_A(x1) = x1 precedence: __ = a____ = isPal = a > a____# = u > o > mark = mark# = and = a__and# = tt = isList = a__isList# = a__isNeList# = a__isList = isNeList = isNePal = a__isNePal# = a__isQid = a__isPal# = nil = a__and = a__isNeList = a__isNePal = a__isPal = e = i = isQid partial status: pi(a____#) = [] pi(__) = [] pi(mark) = [] pi(a____) = [] pi(mark#) = [] pi(and) = [] pi(a__and#) = [] pi(tt) = [] pi(isList) = [] pi(a__isList#) = [] pi(a__isNeList#) = [] pi(a__isList) = [] pi(isNeList) = [] pi(isNePal) = [] pi(a__isNePal#) = [] pi(a__isQid) = [] pi(isPal) = [] pi(a__isPal#) = [1] pi(nil) = [] pi(a__and) = [] pi(a__isNeList) = [1] pi(a__isNePal) = [1] pi(a__isPal) = [] pi(a) = [] pi(e) = [] pi(i) = [] pi(o) = [] pi(u) = [] pi(isQid) = [] The next rules are strictly ordered: p23 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(__(X,Y),Z) -> mark#(X) p3: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p4: a____#(__(X,Y),Z) -> a____#(mark(Y),mark(Z)) p5: a____#(__(X,Y),Z) -> mark#(Y) p6: mark#(__(X1,X2)) -> mark#(X1) p7: mark#(__(X1,X2)) -> mark#(X2) p8: mark#(and(X1,X2)) -> a__and#(mark(X1),X2) p9: a__and#(tt(),X) -> mark#(X) p10: mark#(and(X1,X2)) -> mark#(X1) p11: mark#(isList(X)) -> a__isList#(X) p12: a__isList#(V) -> a__isNeList#(V) p13: a__isNeList#(__(V1,V2)) -> a__and#(a__isList(V1),isNeList(V2)) p14: a__isNeList#(__(V1,V2)) -> a__isList#(V1) p15: a__isList#(__(V1,V2)) -> a__and#(a__isList(V1),isList(V2)) p16: a__isList#(__(V1,V2)) -> a__isList#(V1) p17: a__isNeList#(__(V1,V2)) -> a__isNeList#(V1) p18: mark#(isNeList(X)) -> a__isNeList#(X) p19: mark#(isNePal(X)) -> a__isNePal#(X) p20: a__isNePal#(__(I,__(P,I))) -> a__and#(a__isQid(I),isPal(P)) p21: mark#(isPal(X)) -> a__isPal#(X) p22: a__isPal#(V) -> a__isNePal#(V) p23: a____#(X,nil()) -> mark#(X) p24: a____#(nil(),X) -> mark#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19, p20, p21, p22, p23, p24} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(nil(),X) -> mark#(X) p3: mark#(isPal(X)) -> a__isPal#(X) p4: a__isPal#(V) -> a__isNePal#(V) p5: a__isNePal#(__(I,__(P,I))) -> a__and#(a__isQid(I),isPal(P)) p6: a__and#(tt(),X) -> mark#(X) p7: mark#(isNePal(X)) -> a__isNePal#(X) p8: mark#(isNeList(X)) -> a__isNeList#(X) p9: a__isNeList#(__(V1,V2)) -> a__isNeList#(V1) p10: a__isNeList#(__(V1,V2)) -> a__isList#(V1) p11: a__isList#(__(V1,V2)) -> a__isList#(V1) p12: a__isList#(__(V1,V2)) -> a__and#(a__isList(V1),isList(V2)) p13: a__isList#(V) -> a__isNeList#(V) p14: a__isNeList#(__(V1,V2)) -> a__and#(a__isList(V1),isNeList(V2)) p15: mark#(isList(X)) -> a__isList#(X) p16: mark#(and(X1,X2)) -> mark#(X1) p17: mark#(and(X1,X2)) -> a__and#(mark(X1),X2) p18: mark#(__(X1,X2)) -> mark#(X2) p19: mark#(__(X1,X2)) -> mark#(X1) p20: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p21: a____#(X,nil()) -> mark#(X) p22: a____#(__(X,Y),Z) -> mark#(Y) p23: a____#(__(X,Y),Z) -> a____#(mark(Y),mark(Z)) p24: a____#(__(X,Y),Z) -> mark#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33, r34, r35, r36, r37, r38, r39, r40 Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a____#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (9,3) ___A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (8,4) mark_A(x1) = ((1,0),(0,0)) x1 + (0,13) a_____A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (8,12) nil_A() = (9,0) mark#_A(x1) = ((1,0),(0,0)) x1 + (17,10) isPal_A(x1) = ((1,0),(0,0)) x1 + (3,3) a__isPal#_A(x1) = ((1,0),(0,0)) x1 + (20,7) a__isNePal#_A(x1) = ((1,0),(0,0)) x1 + (20,6) a__and#_A(x1,x2) = ((1,0),(0,0)) x2 + (18,5) a__isQid_A(x1) = ((1,0),(0,0)) x1 + (2,0) tt_A() = (2,13) isNePal_A(x1) = ((1,0),(0,0)) x1 + (3,7) isNeList_A(x1) = ((1,0),(0,0)) x1 + (3,1) a__isNeList#_A(x1) = ((1,0),(0,0)) x1 + (14,9) a__isList#_A(x1) = ((1,0),(0,0)) x1 + (15,6) a__isList_A(x1) = ((1,0),(0,0)) x1 + (4,13) isList_A(x1) = ((1,0),(0,0)) x1 + (4,3) and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,1) a__and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,1) a__isNeList_A(x1) = ((1,0),(0,0)) x1 + (3,2) a__isNePal_A(x1) = ((1,0),(0,0)) x1 + (3,12) a__isPal_A(x1) = ((1,0),(0,0)) x1 + (3,12) a_A() = (3,1) e_A() = (3,12) i_A() = (1,0) o_A() = (3,0) u_A() = (3,0) isQid_A(x1) = ((1,0),(0,0)) x1 + (2,0) precedence: a__isPal > isPal = a__isNeList# > a____ > a____# = mark = mark# = isNeList = a__isList# = a__isList = a__and = a__isNeList = a__isNePal > nil > isList > and > a__isPal# = a__isNePal# = a__and# = isNePal > __ = o > a__isQid = tt = a = i > e = u = isQid partial status: pi(a____#) = [] pi(__) = [] pi(mark) = [] pi(a____) = [] pi(nil) = [] pi(mark#) = [] pi(isPal) = [] pi(a__isPal#) = [] pi(a__isNePal#) = [] pi(a__and#) = [] pi(a__isQid) = [] pi(tt) = [] pi(isNePal) = [] pi(isNeList) = [] pi(a__isNeList#) = [] pi(a__isList#) = [] pi(a__isList) = [] pi(isList) = [] pi(and) = [] pi(a__and) = [] pi(a__isNeList) = [] pi(a__isNePal) = [] pi(a__isPal) = [] pi(a) = [] pi(e) = [] pi(i) = [] pi(o) = [] pi(u) = [] pi(isQid) = [] The next rules are strictly ordered: p6 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(nil(),X) -> mark#(X) p3: mark#(isPal(X)) -> a__isPal#(X) p4: a__isPal#(V) -> a__isNePal#(V) p5: a__isNePal#(__(I,__(P,I))) -> a__and#(a__isQid(I),isPal(P)) p6: mark#(isNePal(X)) -> a__isNePal#(X) p7: mark#(isNeList(X)) -> a__isNeList#(X) p8: a__isNeList#(__(V1,V2)) -> a__isNeList#(V1) p9: a__isNeList#(__(V1,V2)) -> a__isList#(V1) p10: a__isList#(__(V1,V2)) -> a__isList#(V1) p11: a__isList#(__(V1,V2)) -> a__and#(a__isList(V1),isList(V2)) p12: a__isList#(V) -> a__isNeList#(V) p13: a__isNeList#(__(V1,V2)) -> a__and#(a__isList(V1),isNeList(V2)) p14: mark#(isList(X)) -> a__isList#(X) p15: mark#(and(X1,X2)) -> mark#(X1) p16: mark#(and(X1,X2)) -> a__and#(mark(X1),X2) p17: mark#(__(X1,X2)) -> mark#(X2) p18: mark#(__(X1,X2)) -> mark#(X1) p19: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p20: a____#(X,nil()) -> mark#(X) p21: a____#(__(X,Y),Z) -> mark#(Y) p22: a____#(__(X,Y),Z) -> a____#(mark(Y),mark(Z)) p23: a____#(__(X,Y),Z) -> mark#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The estimated dependency graph contains the following SCCs: {p1, p2, p15, p17, p18, p19, p20, p21, p22, p23} {p8, p9, p10, p12} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(__(X,Y),Z) -> mark#(X) p3: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p4: a____#(__(X,Y),Z) -> a____#(mark(Y),mark(Z)) p5: a____#(__(X,Y),Z) -> mark#(Y) p6: mark#(__(X1,X2)) -> mark#(X1) p7: mark#(__(X1,X2)) -> mark#(X2) p8: mark#(and(X1,X2)) -> mark#(X1) p9: a____#(X,nil()) -> mark#(X) p10: a____#(nil(),X) -> mark#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33, r34, r35, r36, r37, r38, r39, r40 Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a____#_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (2,6) ___A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (24,36) mark_A(x1) = x1 + (0,5) a_____A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (24,36) mark#_A(x1) = ((1,0),(1,1)) x1 + (2,6) and_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (3,1) nil_A() = (1,1) a__and_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (3,2) tt_A() = (2,2) a__isList_A(x1) = x1 + (20,6) a__isNeList_A(x1) = x1 + (20,6) isList_A(x1) = x1 + (20,1) a__isQid_A(x1) = ((1,0),(0,0)) x1 + (2,36) isNeList_A(x1) = x1 + (20,1) a__isNePal_A(x1) = ((1,0),(1,0)) x1 + (3,37) isPal_A(x1) = ((1,0),(1,0)) x1 + (18,86) a__isPal_A(x1) = ((1,0),(1,0)) x1 + (18,86) a_A() = (2,3) e_A() = (1,3) i_A() = (1,0) o_A() = (1,1) u_A() = (3,3) isQid_A(x1) = ((1,0),(0,0)) x1 + (2,32) isNePal_A(x1) = ((1,0),(1,0)) x1 + (3,33) precedence: i = o = u > __ = mark = a____ = a__and = a__isList = a__isNeList = a__isNePal = isPal = a__isPal > isNeList > tt = a__isQid = e = isQid > nil = isList > a > a____# = mark# > and > isNePal partial status: pi(a____#) = [1] pi(__) = [] pi(mark) = [] pi(a____) = [] pi(mark#) = [1] pi(and) = [] pi(nil) = [] pi(a__and) = [] pi(tt) = [] pi(a__isList) = [] pi(a__isNeList) = [] pi(isList) = [] pi(a__isQid) = [] pi(isNeList) = [] pi(a__isNePal) = [] pi(isPal) = [] pi(a__isPal) = [] pi(a) = [] pi(e) = [] pi(i) = [] pi(o) = [] pi(u) = [] pi(isQid) = [] pi(isNePal) = [] The next rules are strictly ordered: p10 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(__(X,Y),Z) -> mark#(X) p3: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p4: a____#(__(X,Y),Z) -> a____#(mark(Y),mark(Z)) p5: a____#(__(X,Y),Z) -> mark#(Y) p6: mark#(__(X1,X2)) -> mark#(X1) p7: mark#(__(X1,X2)) -> mark#(X2) p8: mark#(and(X1,X2)) -> mark#(X1) p9: a____#(X,nil()) -> mark#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6, p7, p8, p9} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(X,nil()) -> mark#(X) p3: mark#(and(X1,X2)) -> mark#(X1) p4: mark#(__(X1,X2)) -> mark#(X2) p5: mark#(__(X1,X2)) -> mark#(X1) p6: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p7: a____#(__(X,Y),Z) -> mark#(Y) p8: a____#(__(X,Y),Z) -> a____#(mark(Y),mark(Z)) p9: a____#(__(X,Y),Z) -> mark#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33, r34, r35, r36, r37, r38, r39, r40 Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a____#_A(x1,x2) = ((1,0),(0,0)) x1 + (2,19) ___A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (8,19) mark_A(x1) = ((1,0),(1,1)) x1 + (0,6) a_____A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (8,33) nil_A() = (2,35) mark#_A(x1) = ((1,0),(0,0)) x1 + (1,36) and_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (0,8) a__and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (0,10) tt_A() = (1,34) a__isList_A(x1) = ((1,0),(1,0)) x1 + (7,31) a__isNeList_A(x1) = ((1,0),(1,0)) x1 + (0,2) isList_A(x1) = ((1,0),(0,0)) x1 + (7,20) a__isQid_A(x1) = ((1,0),(1,0)) x1 + (0,1) isNeList_A(x1) = ((1,0),(0,0)) x1 + (0,1) a__isNePal_A(x1) = ((1,0),(0,0)) x1 + (1,19) isPal_A(x1) = x1 + (7,1) a__isPal_A(x1) = x1 + (7,1) a_A() = (34,35) e_A() = (34,1) i_A() = (2,35) o_A() = (34,35) u_A() = (34,35) isQid_A(x1) = ((1,0),(0,0)) x1 isNePal_A(x1) = ((1,0),(0,0)) x1 + (1,13) precedence: mark = and = a__and = a__isList > __ = a____ > a____# = mark# > a__isNePal = isPal = a__isPal > nil = a__isNeList = isList = isNeList > a__isQid > a = e = i = o > tt = u = isQid = isNePal partial status: pi(a____#) = [] pi(__) = [] pi(mark) = [] pi(a____) = [] pi(nil) = [] pi(mark#) = [] pi(and) = [] pi(a__and) = [] pi(tt) = [] pi(a__isList) = [] pi(a__isNeList) = [] pi(isList) = [] pi(a__isQid) = [] pi(isNeList) = [] pi(a__isNePal) = [] pi(isPal) = [] pi(a__isPal) = [] pi(a) = [] pi(e) = [] pi(i) = [] pi(o) = [] pi(u) = [] pi(isQid) = [] pi(isNePal) = [] The next rules are strictly ordered: p8 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(X,nil()) -> mark#(X) p3: mark#(and(X1,X2)) -> mark#(X1) p4: mark#(__(X1,X2)) -> mark#(X2) p5: mark#(__(X1,X2)) -> mark#(X1) p6: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p7: a____#(__(X,Y),Z) -> mark#(Y) p8: a____#(__(X,Y),Z) -> mark#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6, p7, p8} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(__(X,Y),Z) -> mark#(X) p3: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p4: a____#(__(X,Y),Z) -> mark#(Y) p5: mark#(__(X1,X2)) -> mark#(X1) p6: mark#(__(X1,X2)) -> mark#(X2) p7: mark#(and(X1,X2)) -> mark#(X1) p8: a____#(X,nil()) -> mark#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33, r34, r35, r36, r37, r38, r39, r40 Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a____#_A(x1,x2) = ((1,0),(1,1)) x1 + (11,4) ___A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (12,2) mark_A(x1) = ((1,0),(1,1)) x1 a_____A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (12,3) mark#_A(x1) = ((1,0),(1,1)) x1 + (7,3) and_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (7,4) nil_A() = (8,4) a__and_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (7,5) tt_A() = (3,1) a__isList_A(x1) = ((1,0),(0,0)) x1 + (4,21) a__isNeList_A(x1) = ((1,0),(0,0)) x1 + (3,16) isList_A(x1) = ((1,0),(0,0)) x1 + (4,17) a__isQid_A(x1) = ((1,0),(0,0)) x1 + (2,15) isNeList_A(x1) = ((1,0),(0,0)) x1 + (3,14) a__isNePal_A(x1) = ((1,0),(1,0)) x1 + (3,11) isPal_A(x1) = x1 + (12,1) a__isPal_A(x1) = ((1,0),(1,1)) x1 + (12,12) a_A() = (1,0) e_A() = (4,2) i_A() = (4,2) o_A() = (4,2) u_A() = (1,2) isQid_A(x1) = ((1,0),(0,0)) x1 + (2,14) isNePal_A(x1) = ((1,0),(0,0)) x1 + (3,10) precedence: mark = a____ = nil = a__and = a__isList = a__isNeList = isList = a__isQid = isNeList = a__isNePal = isPal = a__isPal = isQid = isNePal > __ > a____# = mark# = and = a = e = i = o = u > tt partial status: pi(a____#) = [1] pi(__) = [1] pi(mark) = [] pi(a____) = [] pi(mark#) = [1] pi(and) = [1] pi(nil) = [] pi(a__and) = [] pi(tt) = [] pi(a__isList) = [] pi(a__isNeList) = [] pi(isList) = [] pi(a__isQid) = [] pi(isNeList) = [] pi(a__isNePal) = [] pi(isPal) = [] pi(a__isPal) = [] pi(a) = [] pi(e) = [] pi(i) = [] pi(o) = [] pi(u) = [] pi(isQid) = [] pi(isNePal) = [] The next rules are strictly ordered: p4 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(__(X,Y),Z) -> mark#(X) p3: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p4: mark#(__(X1,X2)) -> mark#(X1) p5: mark#(__(X1,X2)) -> mark#(X2) p6: mark#(and(X1,X2)) -> mark#(X1) p7: a____#(X,nil()) -> mark#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6, p7} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(X,nil()) -> mark#(X) p3: mark#(and(X1,X2)) -> mark#(X1) p4: mark#(__(X1,X2)) -> mark#(X2) p5: mark#(__(X1,X2)) -> mark#(X1) p6: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p7: a____#(__(X,Y),Z) -> mark#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33, r34, r35, r36, r37, r38, r39, r40 Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a____#_A(x1,x2) = ((0,0),(1,0)) x1 + (2,19) ___A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (11,30) mark_A(x1) = ((1,0),(1,1)) x1 + (0,41) a_____A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (11,41) nil_A() = (3,17) mark#_A(x1) = ((0,0),(1,0)) x1 + (2,18) and_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (2,19) a__and_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (2,20) tt_A() = (1,21) a__isList_A(x1) = x1 + (4,5) a__isNeList_A(x1) = ((1,0),(0,0)) x1 + (3,4) isList_A(x1) = x1 + (4,1) a__isQid_A(x1) = (2,3) isNeList_A(x1) = ((1,0),(0,0)) x1 + (3,3) a__isNePal_A(x1) = ((1,0),(1,1)) x1 + (6,2) isPal_A(x1) = ((1,0),(1,1)) x1 + (23,55) a__isPal_A(x1) = ((1,0),(1,1)) x1 + (23,56) a_A() = (2,22) e_A() = (2,22) i_A() = (2,22) o_A() = (2,22) u_A() = (2,22) isQid_A(x1) = (2,1) isNePal_A(x1) = ((1,0),(1,1)) x1 + (6,1) precedence: __ = mark = a____ = nil = mark# = a__and = tt = a__isNePal = isPal = a__isPal > a____# > and = a__isList = a__isNeList = isList = a__isQid > isNeList = a = i > e = o = u = isQid = isNePal partial status: pi(a____#) = [] pi(__) = [] pi(mark) = [] pi(a____) = [] pi(nil) = [] pi(mark#) = [] pi(and) = [1] pi(a__and) = [] pi(tt) = [] pi(a__isList) = [1] pi(a__isNeList) = [] pi(isList) = [1] pi(a__isQid) = [] pi(isNeList) = [] pi(a__isNePal) = [] pi(isPal) = [] pi(a__isPal) = [] pi(a) = [] pi(e) = [] pi(i) = [] pi(o) = [] pi(u) = [] pi(isQid) = [] pi(isNePal) = [] The next rules are strictly ordered: p2 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: mark#(and(X1,X2)) -> mark#(X1) p3: mark#(__(X1,X2)) -> mark#(X2) p4: mark#(__(X1,X2)) -> mark#(X1) p5: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p6: a____#(__(X,Y),Z) -> mark#(X) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: a____#(__(X,Y),Z) -> mark#(X) p3: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p4: mark#(__(X1,X2)) -> mark#(X1) p5: mark#(__(X1,X2)) -> mark#(X2) p6: mark#(and(X1,X2)) -> mark#(X1) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33, r34, r35, r36, r37, r38, r39, r40 Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a____#_A(x1,x2) = ((1,0),(0,0)) x1 + (1,3) ___A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (13,4) mark_A(x1) = ((1,0),(0,0)) x1 + (0,20) a_____A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (13,19) mark#_A(x1) = ((1,0),(1,1)) x1 + (1,2) and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,1) a__and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,7) tt_A() = (2,1) a__isList_A(x1) = ((1,0),(0,0)) x1 + (6,7) a__isNeList_A(x1) = ((1,0),(0,0)) x1 + (4,6) nil_A() = (1,0) isList_A(x1) = ((1,0),(0,0)) x1 + (6,4) a__isQid_A(x1) = (3,5) isNeList_A(x1) = ((1,0),(0,0)) x1 + (4,4) a__isNePal_A(x1) = ((1,0),(0,0)) x1 + (4,8) isPal_A(x1) = ((1,0),(0,0)) x1 + (13,0) a__isPal_A(x1) = ((1,0),(0,0)) x1 + (13,19) a_A() = (3,2) e_A() = (3,2) i_A() = (3,2) o_A() = (3,2) u_A() = (1,0) isQid_A(x1) = (3,0) isNePal_A(x1) = ((1,0),(0,0)) x1 + (4,0) precedence: __ = mark = a____ = mark# = a__and = a__isQid = a__isNePal = a__isPal = u = isNePal > a__isList = isPal > a__isNeList = isNeList > isList > a = e = i = o > tt > a____# = and = nil = isQid partial status: pi(a____#) = [] pi(__) = [] pi(mark) = [] pi(a____) = [] pi(mark#) = [1] pi(and) = [] pi(a__and) = [] pi(tt) = [] pi(a__isList) = [] pi(a__isNeList) = [] pi(nil) = [] pi(isList) = [] pi(a__isQid) = [] pi(isNeList) = [] pi(a__isNePal) = [] pi(isPal) = [] pi(a__isPal) = [] pi(a) = [] pi(e) = [] pi(i) = [] pi(o) = [] pi(u) = [] pi(isQid) = [] pi(isNePal) = [] The next rules are strictly ordered: p2 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) p2: mark#(__(X1,X2)) -> a____#(mark(X1),mark(X2)) p3: mark#(__(X1,X2)) -> mark#(X1) p4: mark#(__(X1,X2)) -> mark#(X2) p5: mark#(and(X1,X2)) -> mark#(X1) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The estimated dependency graph contains the following SCCs: {p3, p4, p5} {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: mark#(and(X1,X2)) -> mark#(X1) p2: mark#(__(X1,X2)) -> mark#(X2) p3: mark#(__(X1,X2)) -> mark#(X1) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of (no rules) Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: mark#_A(x1) = ((1,0),(0,0)) x1 + (1,1) and_A(x1,x2) = x1 + x2 + (2,2) ___A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (1,2) precedence: mark# = and = __ partial status: pi(mark#) = [] pi(and) = [2] pi(__) = [2] The next rules are strictly ordered: p2 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: mark#(and(X1,X2)) -> mark#(X1) p2: mark#(__(X1,X2)) -> mark#(X1) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: mark#(and(X1,X2)) -> mark#(X1) p2: mark#(__(X1,X2)) -> mark#(X1) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of (no rules) Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: mark#_A(x1) = ((1,0),(0,0)) x1 and_A(x1,x2) = ((1,0),(0,0)) x1 + x2 ___A(x1,x2) = ((1,0),(0,0)) x1 + (1,1) precedence: and = __ > mark# partial status: pi(mark#) = [] pi(and) = [2] pi(__) = [] The next rules are strictly ordered: p2 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: mark#(and(X1,X2)) -> mark#(X1) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: mark#(and(X1,X2)) -> mark#(X1) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of (no rules) Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: mark#_A(x1) = ((1,0),(1,1)) x1 and_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,1) precedence: mark# = and partial status: pi(mark#) = [1] pi(and) = [1] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a____#(__(X,Y),Z) -> a____#(mark(X),a____(mark(Y),mark(Z))) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33, r34, r35, r36, r37, r38, r39, r40 Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a____#_A(x1,x2) = x1 + (1,1) ___A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (8,5) mark_A(x1) = ((1,0),(1,1)) x1 + (0,3) a_____A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (8,6) a__and_A(x1,x2) = ((1,0),(0,0)) x2 + (4,6) tt_A() = (1,2) a__isList_A(x1) = ((1,0),(0,0)) x1 + (7,9) a__isNeList_A(x1) = ((1,0),(0,0)) x1 + (3,7) nil_A() = (2,1) isList_A(x1) = ((1,0),(0,0)) x1 + (7,8) a__isQid_A(x1) = (2,3) isNeList_A(x1) = ((1,0),(0,0)) x1 + (3,6) a__isNePal_A(x1) = ((1,0),(0,0)) x1 + (3,7) isPal_A(x1) = ((1,0),(0,0)) x1 + (5,4) a__isPal_A(x1) = ((1,0),(0,0)) x1 + (5,8) a_A() = (2,3) e_A() = (2,3) i_A() = (2,3) o_A() = (2,3) u_A() = (2,3) and_A(x1,x2) = ((1,0),(0,0)) x2 + (4,1) isQid_A(x1) = (2,1) isNePal_A(x1) = ((1,0),(0,0)) x1 + (3,2) precedence: a____# > __ = mark = a____ = a__and = a__isList = a__isNeList = a__isQid = a__isNePal = and > isPal = a__isPal > tt = nil = isList = isNeList = a = e = i = o = u = isNePal > isQid partial status: pi(a____#) = [1] pi(__) = [] pi(mark) = [] pi(a____) = [] pi(a__and) = [] pi(tt) = [] pi(a__isList) = [] pi(a__isNeList) = [] pi(nil) = [] pi(isList) = [] pi(a__isQid) = [] pi(isNeList) = [] pi(a__isNePal) = [] pi(isPal) = [] pi(a__isPal) = [] pi(a) = [] pi(e) = [] pi(i) = [] pi(o) = [] pi(u) = [] pi(and) = [] pi(isQid) = [] pi(isNePal) = [] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a__isNeList#(__(V1,V2)) -> a__isNeList#(V1) p2: a__isNeList#(__(V1,V2)) -> a__isList#(V1) p3: a__isList#(V) -> a__isNeList#(V) p4: a__isList#(__(V1,V2)) -> a__isList#(V1) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of (no rules) Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a__isNeList#_A(x1) = ((1,0),(1,1)) x1 + (1,3) ___A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (3,2) a__isList#_A(x1) = ((1,0),(1,0)) x1 + (2,1) precedence: a__isNeList# = __ = a__isList# partial status: pi(a__isNeList#) = [1] pi(__) = [2] pi(a__isList#) = [] The next rules are strictly ordered: p1 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a__isNeList#(__(V1,V2)) -> a__isList#(V1) p2: a__isList#(V) -> a__isNeList#(V) p3: a__isList#(__(V1,V2)) -> a__isList#(V1) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The estimated dependency graph contains the following SCCs: {p1, p2, p3} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a__isNeList#(__(V1,V2)) -> a__isList#(V1) p2: a__isList#(__(V1,V2)) -> a__isList#(V1) p3: a__isList#(V) -> a__isNeList#(V) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of (no rules) Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a__isNeList#_A(x1) = ((0,0),(1,0)) x1 ___A(x1,x2) = ((1,0),(1,1)) x1 + (2,2) a__isList#_A(x1) = ((0,0),(1,0)) x1 + (0,1) precedence: __ > a__isNeList# = a__isList# partial status: pi(a__isNeList#) = [] pi(__) = [] pi(a__isList#) = [] The next rules are strictly ordered: p2 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a__isNeList#(__(V1,V2)) -> a__isList#(V1) p2: a__isList#(V) -> a__isNeList#(V) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a__isNeList#(__(V1,V2)) -> a__isList#(V1) p2: a__isList#(V) -> a__isNeList#(V) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The set of usable rules consists of (no rules) Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a__isNeList#_A(x1) = x1 ___A(x1,x2) = x1 + x2 + (2,2) a__isList#_A(x1) = x1 + (1,1) precedence: __ > a__isNeList# = a__isList# partial status: pi(a__isNeList#) = [1] pi(__) = [2] pi(a__isList#) = [1] The next rules are strictly ordered: p1 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a__isList#(V) -> a__isNeList#(V) and R consists of: r1: a____(__(X,Y),Z) -> a____(mark(X),a____(mark(Y),mark(Z))) r2: a____(X,nil()) -> mark(X) r3: a____(nil(),X) -> mark(X) r4: a__and(tt(),X) -> mark(X) r5: a__isList(V) -> a__isNeList(V) r6: a__isList(nil()) -> tt() r7: a__isList(__(V1,V2)) -> a__and(a__isList(V1),isList(V2)) r8: a__isNeList(V) -> a__isQid(V) r9: a__isNeList(__(V1,V2)) -> a__and(a__isList(V1),isNeList(V2)) r10: a__isNeList(__(V1,V2)) -> a__and(a__isNeList(V1),isList(V2)) r11: a__isNePal(V) -> a__isQid(V) r12: a__isNePal(__(I,__(P,I))) -> a__and(a__isQid(I),isPal(P)) r13: a__isPal(V) -> a__isNePal(V) r14: a__isPal(nil()) -> tt() r15: a__isQid(a()) -> tt() r16: a__isQid(e()) -> tt() r17: a__isQid(i()) -> tt() r18: a__isQid(o()) -> tt() r19: a__isQid(u()) -> tt() r20: mark(__(X1,X2)) -> a____(mark(X1),mark(X2)) r21: mark(and(X1,X2)) -> a__and(mark(X1),X2) r22: mark(isList(X)) -> a__isList(X) r23: mark(isNeList(X)) -> a__isNeList(X) r24: mark(isQid(X)) -> a__isQid(X) r25: mark(isNePal(X)) -> a__isNePal(X) r26: mark(isPal(X)) -> a__isPal(X) r27: mark(nil()) -> nil() r28: mark(tt()) -> tt() r29: mark(a()) -> a() r30: mark(e()) -> e() r31: mark(i()) -> i() r32: mark(o()) -> o() r33: mark(u()) -> u() r34: a____(X1,X2) -> __(X1,X2) r35: a__and(X1,X2) -> and(X1,X2) r36: a__isList(X) -> isList(X) r37: a__isNeList(X) -> isNeList(X) r38: a__isQid(X) -> isQid(X) r39: a__isNePal(X) -> isNePal(X) r40: a__isPal(X) -> isPal(X) The estimated dependency graph contains the following SCCs: (no SCCs)