YES We show the termination of the TRS R: a__incr(nil()) -> nil() a__incr(cons(X,L)) -> cons(s(mark(X)),incr(L)) a__adx(nil()) -> nil() a__adx(cons(X,L)) -> a__incr(cons(mark(X),adx(L))) a__nats() -> a__adx(a__zeros()) a__zeros() -> cons(|0|(),zeros()) a__head(cons(X,L)) -> mark(X) a__tail(cons(X,L)) -> mark(L) mark(incr(X)) -> a__incr(mark(X)) mark(adx(X)) -> a__adx(mark(X)) mark(nats()) -> a__nats() mark(zeros()) -> a__zeros() mark(head(X)) -> a__head(mark(X)) mark(tail(X)) -> a__tail(mark(X)) mark(nil()) -> nil() mark(cons(X1,X2)) -> cons(mark(X1),X2) mark(s(X)) -> s(mark(X)) mark(|0|()) -> |0|() a__incr(X) -> incr(X) a__adx(X) -> adx(X) a__nats() -> nats() a__zeros() -> zeros() a__head(X) -> head(X) a__tail(X) -> tail(X) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a__incr#(cons(X,L)) -> mark#(X) p2: a__adx#(cons(X,L)) -> a__incr#(cons(mark(X),adx(L))) p3: a__adx#(cons(X,L)) -> mark#(X) p4: a__nats#() -> a__adx#(a__zeros()) p5: a__nats#() -> a__zeros#() p6: a__head#(cons(X,L)) -> mark#(X) p7: a__tail#(cons(X,L)) -> mark#(L) p8: mark#(incr(X)) -> a__incr#(mark(X)) p9: mark#(incr(X)) -> mark#(X) p10: mark#(adx(X)) -> a__adx#(mark(X)) p11: mark#(adx(X)) -> mark#(X) p12: mark#(nats()) -> a__nats#() p13: mark#(zeros()) -> a__zeros#() p14: mark#(head(X)) -> a__head#(mark(X)) p15: mark#(head(X)) -> mark#(X) p16: mark#(tail(X)) -> a__tail#(mark(X)) p17: mark#(tail(X)) -> mark#(X) p18: mark#(cons(X1,X2)) -> mark#(X1) p19: mark#(s(X)) -> mark#(X) and R consists of: r1: a__incr(nil()) -> nil() r2: a__incr(cons(X,L)) -> cons(s(mark(X)),incr(L)) r3: a__adx(nil()) -> nil() r4: a__adx(cons(X,L)) -> a__incr(cons(mark(X),adx(L))) r5: a__nats() -> a__adx(a__zeros()) r6: a__zeros() -> cons(|0|(),zeros()) r7: a__head(cons(X,L)) -> mark(X) r8: a__tail(cons(X,L)) -> mark(L) r9: mark(incr(X)) -> a__incr(mark(X)) r10: mark(adx(X)) -> a__adx(mark(X)) r11: mark(nats()) -> a__nats() r12: mark(zeros()) -> a__zeros() r13: mark(head(X)) -> a__head(mark(X)) r14: mark(tail(X)) -> a__tail(mark(X)) r15: mark(nil()) -> nil() r16: mark(cons(X1,X2)) -> cons(mark(X1),X2) r17: mark(s(X)) -> s(mark(X)) r18: mark(|0|()) -> |0|() r19: a__incr(X) -> incr(X) r20: a__adx(X) -> adx(X) r21: a__nats() -> nats() r22: a__zeros() -> zeros() r23: a__head(X) -> head(X) r24: a__tail(X) -> tail(X) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p6, p7, p8, p9, p10, p11, p12, p14, p15, p16, p17, p18, p19} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a__incr#(cons(X,L)) -> mark#(X) p2: mark#(s(X)) -> mark#(X) p3: mark#(cons(X1,X2)) -> mark#(X1) p4: mark#(tail(X)) -> mark#(X) p5: mark#(tail(X)) -> a__tail#(mark(X)) p6: a__tail#(cons(X,L)) -> mark#(L) p7: mark#(head(X)) -> mark#(X) p8: mark#(head(X)) -> a__head#(mark(X)) p9: a__head#(cons(X,L)) -> mark#(X) p10: mark#(nats()) -> a__nats#() p11: a__nats#() -> a__adx#(a__zeros()) p12: a__adx#(cons(X,L)) -> mark#(X) p13: mark#(adx(X)) -> mark#(X) p14: mark#(adx(X)) -> a__adx#(mark(X)) p15: a__adx#(cons(X,L)) -> a__incr#(cons(mark(X),adx(L))) p16: mark#(incr(X)) -> mark#(X) p17: mark#(incr(X)) -> a__incr#(mark(X)) and R consists of: r1: a__incr(nil()) -> nil() r2: a__incr(cons(X,L)) -> cons(s(mark(X)),incr(L)) r3: a__adx(nil()) -> nil() r4: a__adx(cons(X,L)) -> a__incr(cons(mark(X),adx(L))) r5: a__nats() -> a__adx(a__zeros()) r6: a__zeros() -> cons(|0|(),zeros()) r7: a__head(cons(X,L)) -> mark(X) r8: a__tail(cons(X,L)) -> mark(L) r9: mark(incr(X)) -> a__incr(mark(X)) r10: mark(adx(X)) -> a__adx(mark(X)) r11: mark(nats()) -> a__nats() r12: mark(zeros()) -> a__zeros() r13: mark(head(X)) -> a__head(mark(X)) r14: mark(tail(X)) -> a__tail(mark(X)) r15: mark(nil()) -> nil() r16: mark(cons(X1,X2)) -> cons(mark(X1),X2) r17: mark(s(X)) -> s(mark(X)) r18: mark(|0|()) -> |0|() r19: a__incr(X) -> incr(X) r20: a__adx(X) -> adx(X) r21: a__nats() -> nats() r22: a__zeros() -> zeros() r23: a__head(X) -> head(X) r24: a__tail(X) -> tail(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 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a__incr#_A(x1) = ((1,0),(0,0)) x1 + (0,2) cons_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (0,3) mark#_A(x1) = ((1,0),(0,0)) x1 + (0,2) s_A(x1) = x1 tail_A(x1) = x1 + (1,0) a__tail#_A(x1) = x1 mark_A(x1) = x1 head_A(x1) = x1 + (2,1) a__head#_A(x1) = x1 + (1,2) nats_A() = (3,8) a__nats#_A() = (3,2) a__adx#_A(x1) = ((1,0),(0,0)) x1 + (0,2) a__zeros_A() = (2,7) adx_A(x1) = ((1,0),(1,1)) x1 incr_A(x1) = x1 a__incr_A(x1) = x1 nil_A() = (1,0) a__adx_A(x1) = ((1,0),(1,1)) x1 a__nats_A() = (3,8) a__head_A(x1) = x1 + (2,1) a__tail_A(x1) = x1 + (1,0) |0|_A() = (0,1) zeros_A() = (2,7) precedence: mark = a__zeros = a__head = |0| > zeros > a__tail > s > a__adx > cons = adx = a__incr > tail > head = nats = a__nats > a__incr# = mark# = a__nats# = a__adx# = incr > a__tail# > a__head# = nil partial status: pi(a__incr#) = [] pi(cons) = [1] pi(mark#) = [] pi(s) = [1] pi(tail) = [1] pi(a__tail#) = [1] pi(mark) = [1] pi(head) = [] pi(a__head#) = [1] pi(nats) = [] pi(a__nats#) = [] pi(a__adx#) = [] pi(a__zeros) = [] pi(adx) = [1] pi(incr) = [1] pi(a__incr) = [1] pi(nil) = [] pi(a__adx) = [1] pi(a__nats) = [] pi(a__head) = [1] pi(a__tail) = [1] pi(|0|) = [] pi(zeros) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a__incr#_A(x1) = (2,24) cons_A(x1,x2) = x1 + (9,10) mark#_A(x1) = (2,24) s_A(x1) = ((1,0),(0,0)) x1 + (3,16) tail_A(x1) = (1,12) a__tail#_A(x1) = (3,23) mark_A(x1) = ((0,0),(1,0)) x1 + (5,12) head_A(x1) = (3,0) a__head#_A(x1) = ((0,0),(1,0)) x1 + (2,19) nats_A() = (3,25) a__nats#_A() = (2,24) a__adx#_A(x1) = (2,24) a__zeros_A() = (11,11) adx_A(x1) = ((1,0),(0,0)) x1 + (19,23) incr_A(x1) = ((1,0),(0,0)) x1 + (1,9) a__incr_A(x1) = (18,10) nil_A() = (1,0) a__adx_A(x1) = (19,23) a__nats_A() = (4,14) a__head_A(x1) = (0,0) a__tail_A(x1) = ((0,0),(1,0)) x1 + (2,8) |0|_A() = (1,0) zeros_A() = (0,11) precedence: |0| > zeros > a__adx > nil > mark = head = a__head = a__tail > adx > s > a__zeros > cons = a__incr = a__nats > incr > a__incr# = mark# = a__tail# = a__head# = a__nats# = a__adx# > nats > tail partial status: pi(a__incr#) = [] pi(cons) = [] pi(mark#) = [] pi(s) = [] pi(tail) = [] pi(a__tail#) = [] pi(mark) = [] pi(head) = [] pi(a__head#) = [] pi(nats) = [] pi(a__nats#) = [] pi(a__adx#) = [] pi(a__zeros) = [] pi(adx) = [] pi(incr) = [] pi(a__incr) = [] pi(nil) = [] pi(a__adx) = [] pi(a__nats) = [] pi(a__head) = [] pi(a__tail) = [] pi(|0|) = [] pi(zeros) = [] The next rules are strictly ordered: p4, p5, p6 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a__incr#(cons(X,L)) -> mark#(X) p2: mark#(s(X)) -> mark#(X) p3: mark#(cons(X1,X2)) -> mark#(X1) p4: mark#(head(X)) -> mark#(X) p5: mark#(head(X)) -> a__head#(mark(X)) p6: a__head#(cons(X,L)) -> mark#(X) p7: mark#(nats()) -> a__nats#() p8: a__nats#() -> a__adx#(a__zeros()) p9: a__adx#(cons(X,L)) -> mark#(X) p10: mark#(adx(X)) -> mark#(X) p11: mark#(adx(X)) -> a__adx#(mark(X)) p12: a__adx#(cons(X,L)) -> a__incr#(cons(mark(X),adx(L))) p13: mark#(incr(X)) -> mark#(X) p14: mark#(incr(X)) -> a__incr#(mark(X)) and R consists of: r1: a__incr(nil()) -> nil() r2: a__incr(cons(X,L)) -> cons(s(mark(X)),incr(L)) r3: a__adx(nil()) -> nil() r4: a__adx(cons(X,L)) -> a__incr(cons(mark(X),adx(L))) r5: a__nats() -> a__adx(a__zeros()) r6: a__zeros() -> cons(|0|(),zeros()) r7: a__head(cons(X,L)) -> mark(X) r8: a__tail(cons(X,L)) -> mark(L) r9: mark(incr(X)) -> a__incr(mark(X)) r10: mark(adx(X)) -> a__adx(mark(X)) r11: mark(nats()) -> a__nats() r12: mark(zeros()) -> a__zeros() r13: mark(head(X)) -> a__head(mark(X)) r14: mark(tail(X)) -> a__tail(mark(X)) r15: mark(nil()) -> nil() r16: mark(cons(X1,X2)) -> cons(mark(X1),X2) r17: mark(s(X)) -> s(mark(X)) r18: mark(|0|()) -> |0|() r19: a__incr(X) -> incr(X) r20: a__adx(X) -> adx(X) r21: a__nats() -> nats() r22: a__zeros() -> zeros() r23: a__head(X) -> head(X) r24: a__tail(X) -> tail(X) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a__incr#(cons(X,L)) -> mark#(X) p2: mark#(incr(X)) -> a__incr#(mark(X)) p3: mark#(incr(X)) -> mark#(X) p4: mark#(adx(X)) -> a__adx#(mark(X)) p5: a__adx#(cons(X,L)) -> a__incr#(cons(mark(X),adx(L))) p6: a__adx#(cons(X,L)) -> mark#(X) p7: mark#(adx(X)) -> mark#(X) p8: mark#(nats()) -> a__nats#() p9: a__nats#() -> a__adx#(a__zeros()) p10: mark#(head(X)) -> a__head#(mark(X)) p11: a__head#(cons(X,L)) -> mark#(X) p12: mark#(head(X)) -> mark#(X) p13: mark#(cons(X1,X2)) -> mark#(X1) p14: mark#(s(X)) -> mark#(X) and R consists of: r1: a__incr(nil()) -> nil() r2: a__incr(cons(X,L)) -> cons(s(mark(X)),incr(L)) r3: a__adx(nil()) -> nil() r4: a__adx(cons(X,L)) -> a__incr(cons(mark(X),adx(L))) r5: a__nats() -> a__adx(a__zeros()) r6: a__zeros() -> cons(|0|(),zeros()) r7: a__head(cons(X,L)) -> mark(X) r8: a__tail(cons(X,L)) -> mark(L) r9: mark(incr(X)) -> a__incr(mark(X)) r10: mark(adx(X)) -> a__adx(mark(X)) r11: mark(nats()) -> a__nats() r12: mark(zeros()) -> a__zeros() r13: mark(head(X)) -> a__head(mark(X)) r14: mark(tail(X)) -> a__tail(mark(X)) r15: mark(nil()) -> nil() r16: mark(cons(X1,X2)) -> cons(mark(X1),X2) r17: mark(s(X)) -> s(mark(X)) r18: mark(|0|()) -> |0|() r19: a__incr(X) -> incr(X) r20: a__adx(X) -> adx(X) r21: a__nats() -> nats() r22: a__zeros() -> zeros() r23: a__head(X) -> head(X) r24: a__tail(X) -> tail(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 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a__incr#_A(x1) = x1 + (0,1) cons_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (0,2) mark#_A(x1) = ((1,0),(1,1)) x1 incr_A(x1) = ((1,0),(1,1)) x1 + (0,3) mark_A(x1) = x1 + (0,2) adx_A(x1) = ((1,0),(1,1)) x1 + (1,7) a__adx#_A(x1) = x1 + (1,4) nats_A() = (4,12) a__nats#_A() = (3,13) a__zeros_A() = (1,4) head_A(x1) = ((1,0),(0,0)) x1 + (2,1) a__head#_A(x1) = ((1,0),(0,0)) x1 + (1,0) s_A(x1) = x1 + (0,1) a__incr_A(x1) = ((1,0),(1,1)) x1 + (0,3) nil_A() = (1,1) a__adx_A(x1) = ((1,0),(1,1)) x1 + (1,7) a__nats_A() = (4,13) a__head_A(x1) = ((1,0),(0,0)) x1 + (2,2) a__tail_A(x1) = ((1,0),(0,0)) x1 + (1,2) tail_A(x1) = ((1,0),(0,0)) x1 + (1,1) |0|_A() = (0,1) zeros_A() = (1,4) precedence: mark# = mark = a__nats# = a__head# = a__head = a__tail = |0| > a__adx# > a__adx > adx > head > a__zeros = a__nats > nats > a__incr > incr > a__incr# > cons > nil > s = tail = zeros partial status: pi(a__incr#) = [1] pi(cons) = [1] pi(mark#) = [1] pi(incr) = [1] pi(mark) = [] pi(adx) = [1] pi(a__adx#) = [1] pi(nats) = [] pi(a__nats#) = [] pi(a__zeros) = [] pi(head) = [] pi(a__head#) = [] pi(s) = [1] pi(a__incr) = [1] pi(nil) = [] pi(a__adx) = [1] pi(a__nats) = [] pi(a__head) = [] pi(a__tail) = [] pi(tail) = [] pi(|0|) = [] pi(zeros) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: a__incr#_A(x1) = x1 + (11,4) cons_A(x1,x2) = ((0,0),(1,0)) x1 + (1,4) mark#_A(x1) = ((0,0),(1,0)) x1 + (12,3) incr_A(x1) = ((1,0),(0,0)) x1 + (3,2) mark_A(x1) = (4,2) adx_A(x1) = ((1,0),(0,0)) x1 + (10,1) a__adx#_A(x1) = (12,13) nats_A() = (11,1) a__nats#_A() = (12,13) a__zeros_A() = (2,0) head_A(x1) = (3,1) a__head#_A(x1) = (13,6) s_A(x1) = ((1,0),(0,0)) x1 + (1,0) a__incr_A(x1) = (2,0) nil_A() = (3,2) a__adx_A(x1) = ((1,0),(0,0)) x1 + (11,2) a__nats_A() = (3,1) a__head_A(x1) = (4,2) a__tail_A(x1) = (4,2) tail_A(x1) = (3,2) |0|_A() = (5,3) zeros_A() = (0,1) precedence: |0| > zeros > tail > a__zeros > a__nats# > nats > a__incr# > mark# = a__adx# = a__head# > head > cons > mark = a__head = a__tail > a__nats > s > a__incr = nil = a__adx > adx > incr partial status: pi(a__incr#) = [] pi(cons) = [] pi(mark#) = [] pi(incr) = [] pi(mark) = [] pi(adx) = [] pi(a__adx#) = [] pi(nats) = [] pi(a__nats#) = [] pi(a__zeros) = [] pi(head) = [] pi(a__head#) = [] pi(s) = [] pi(a__incr) = [] pi(nil) = [] pi(a__adx) = [] pi(a__nats) = [] pi(a__head) = [] pi(a__tail) = [] pi(tail) = [] pi(|0|) = [] pi(zeros) = [] The next rules are strictly ordered: p1, p2, p3, p6, p7, p10, p12, p13, p14 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: mark#(adx(X)) -> a__adx#(mark(X)) p2: a__adx#(cons(X,L)) -> a__incr#(cons(mark(X),adx(L))) p3: mark#(nats()) -> a__nats#() p4: a__nats#() -> a__adx#(a__zeros()) p5: a__head#(cons(X,L)) -> mark#(X) and R consists of: r1: a__incr(nil()) -> nil() r2: a__incr(cons(X,L)) -> cons(s(mark(X)),incr(L)) r3: a__adx(nil()) -> nil() r4: a__adx(cons(X,L)) -> a__incr(cons(mark(X),adx(L))) r5: a__nats() -> a__adx(a__zeros()) r6: a__zeros() -> cons(|0|(),zeros()) r7: a__head(cons(X,L)) -> mark(X) r8: a__tail(cons(X,L)) -> mark(L) r9: mark(incr(X)) -> a__incr(mark(X)) r10: mark(adx(X)) -> a__adx(mark(X)) r11: mark(nats()) -> a__nats() r12: mark(zeros()) -> a__zeros() r13: mark(head(X)) -> a__head(mark(X)) r14: mark(tail(X)) -> a__tail(mark(X)) r15: mark(nil()) -> nil() r16: mark(cons(X1,X2)) -> cons(mark(X1),X2) r17: mark(s(X)) -> s(mark(X)) r18: mark(|0|()) -> |0|() r19: a__incr(X) -> incr(X) r20: a__adx(X) -> adx(X) r21: a__nats() -> nats() r22: a__zeros() -> zeros() r23: a__head(X) -> head(X) r24: a__tail(X) -> tail(X) The estimated dependency graph contains the following SCCs: (no SCCs)