YES We show the termination of the TRS R: active(f(x)) -> mark(x) top(active(c())) -> top(mark(c())) top(mark(x)) -> top(check(x)) check(f(x)) -> f(check(x)) check(x) -> start(match(f(X()),x)) match(f(x),f(y)) -> f(match(x,y)) match(X(),x) -> proper(x) proper(c()) -> ok(c()) proper(f(x)) -> f(proper(x)) f(ok(x)) -> ok(f(x)) start(ok(x)) -> found(x) f(found(x)) -> found(f(x)) top(found(x)) -> top(active(x)) active(f(x)) -> f(active(x)) f(mark(x)) -> mark(f(x)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: top#(active(c())) -> top#(mark(c())) p2: top#(mark(x)) -> top#(check(x)) p3: top#(mark(x)) -> check#(x) p4: check#(f(x)) -> f#(check(x)) p5: check#(f(x)) -> check#(x) p6: check#(x) -> start#(match(f(X()),x)) p7: check#(x) -> match#(f(X()),x) p8: check#(x) -> f#(X()) p9: match#(f(x),f(y)) -> f#(match(x,y)) p10: match#(f(x),f(y)) -> match#(x,y) p11: match#(X(),x) -> proper#(x) p12: proper#(f(x)) -> f#(proper(x)) p13: proper#(f(x)) -> proper#(x) p14: f#(ok(x)) -> f#(x) p15: f#(found(x)) -> f#(x) p16: top#(found(x)) -> top#(active(x)) p17: top#(found(x)) -> active#(x) p18: active#(f(x)) -> f#(active(x)) p19: active#(f(x)) -> active#(x) p20: f#(mark(x)) -> f#(x) and R consists of: r1: active(f(x)) -> mark(x) r2: top(active(c())) -> top(mark(c())) r3: top(mark(x)) -> top(check(x)) r4: check(f(x)) -> f(check(x)) r5: check(x) -> start(match(f(X()),x)) r6: match(f(x),f(y)) -> f(match(x,y)) r7: match(X(),x) -> proper(x) r8: proper(c()) -> ok(c()) r9: proper(f(x)) -> f(proper(x)) r10: f(ok(x)) -> ok(f(x)) r11: start(ok(x)) -> found(x) r12: f(found(x)) -> found(f(x)) r13: top(found(x)) -> top(active(x)) r14: active(f(x)) -> f(active(x)) r15: f(mark(x)) -> mark(f(x)) The estimated dependency graph contains the following SCCs: {p1, p2, p16} {p5} {p19} {p10} {p13} {p14, p15, p20} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: top#(active(c())) -> top#(mark(c())) p2: top#(mark(x)) -> top#(check(x)) p3: top#(found(x)) -> top#(active(x)) and R consists of: r1: active(f(x)) -> mark(x) r2: top(active(c())) -> top(mark(c())) r3: top(mark(x)) -> top(check(x)) r4: check(f(x)) -> f(check(x)) r5: check(x) -> start(match(f(X()),x)) r6: match(f(x),f(y)) -> f(match(x,y)) r7: match(X(),x) -> proper(x) r8: proper(c()) -> ok(c()) r9: proper(f(x)) -> f(proper(x)) r10: f(ok(x)) -> ok(f(x)) r11: start(ok(x)) -> found(x) r12: f(found(x)) -> found(f(x)) r13: top(found(x)) -> top(active(x)) r14: active(f(x)) -> f(active(x)) r15: f(mark(x)) -> mark(f(x)) The set of usable rules consists of r1, r4, r5, r6, r7, r8, r9, r10, r11, r12, r14, r15 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: top#_A(x1) = ((1,0),(0,0)) x1 + (1,2) active_A(x1) = x1 + (0,3) c_A() = (2,1) mark_A(x1) = (0,0) check_A(x1) = (0,0) found_A(x1) = ((1,0),(0,0)) x1 proper_A(x1) = (3,2) ok_A(x1) = ((1,0),(0,0)) x1 f_A(x1) = (0,0) match_A(x1,x2) = ((1,0),(0,0)) x1 + (0,3) X_A() = (4,3) start_A(x1) = ((1,0),(0,0)) x1 precedence: check > f > top# = mark > active > c = match = X = start > proper > found = ok partial status: pi(top#) = [] pi(active) = [1] pi(c) = [] pi(mark) = [] pi(check) = [] pi(found) = [] pi(proper) = [] pi(ok) = [] pi(f) = [] pi(match) = [] pi(X) = [] pi(start) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: top#_A(x1) = (2,2) active_A(x1) = x1 + (3,5) c_A() = (0,1) mark_A(x1) = (1,0) check_A(x1) = (6,5) found_A(x1) = (4,3) proper_A(x1) = (3,2) ok_A(x1) = (1,0) f_A(x1) = (5,4) match_A(x1,x2) = (4,6) X_A() = (3,3) start_A(x1) = (5,4) precedence: check > proper = ok = f = match > active = mark > start > found > top# = c > X partial status: pi(top#) = [] pi(active) = [1] pi(c) = [] pi(mark) = [] pi(check) = [] pi(found) = [] pi(proper) = [] pi(ok) = [] pi(f) = [] pi(match) = [] pi(X) = [] pi(start) = [] 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: top#(mark(x)) -> top#(check(x)) p2: top#(found(x)) -> top#(active(x)) and R consists of: r1: active(f(x)) -> mark(x) r2: top(active(c())) -> top(mark(c())) r3: top(mark(x)) -> top(check(x)) r4: check(f(x)) -> f(check(x)) r5: check(x) -> start(match(f(X()),x)) r6: match(f(x),f(y)) -> f(match(x,y)) r7: match(X(),x) -> proper(x) r8: proper(c()) -> ok(c()) r9: proper(f(x)) -> f(proper(x)) r10: f(ok(x)) -> ok(f(x)) r11: start(ok(x)) -> found(x) r12: f(found(x)) -> found(f(x)) r13: top(found(x)) -> top(active(x)) r14: active(f(x)) -> f(active(x)) r15: f(mark(x)) -> mark(f(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: top#(mark(x)) -> top#(check(x)) p2: top#(found(x)) -> top#(active(x)) and R consists of: r1: active(f(x)) -> mark(x) r2: top(active(c())) -> top(mark(c())) r3: top(mark(x)) -> top(check(x)) r4: check(f(x)) -> f(check(x)) r5: check(x) -> start(match(f(X()),x)) r6: match(f(x),f(y)) -> f(match(x,y)) r7: match(X(),x) -> proper(x) r8: proper(c()) -> ok(c()) r9: proper(f(x)) -> f(proper(x)) r10: f(ok(x)) -> ok(f(x)) r11: start(ok(x)) -> found(x) r12: f(found(x)) -> found(f(x)) r13: top(found(x)) -> top(active(x)) r14: active(f(x)) -> f(active(x)) r15: f(mark(x)) -> mark(f(x)) The set of usable rules consists of r1, r4, r5, r6, r7, r8, r9, r10, r11, r12, r14, r15 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: top#_A(x1) = ((0,0),(1,0)) x1 + (2,6) mark_A(x1) = ((1,0),(0,0)) x1 + (3,1) check_A(x1) = ((1,0),(1,1)) x1 + (1,5) found_A(x1) = ((1,0),(0,0)) x1 + (0,3) active_A(x1) = ((1,0),(1,0)) x1 + (0,5) proper_A(x1) = ((1,0),(1,0)) x1 + (0,1) c_A() = (2,1) ok_A(x1) = ((1,0),(0,0)) x1 + (0,2) f_A(x1) = x1 + (5,4) match_A(x1,x2) = ((1,0),(1,1)) x2 + (0,2) X_A() = (2,2) start_A(x1) = x1 + (1,2) precedence: active = proper = match > check = f > X > top# = mark = found = ok = start > c partial status: pi(top#) = [] pi(mark) = [] pi(check) = [1] pi(found) = [] pi(active) = [] pi(proper) = [] pi(c) = [] pi(ok) = [] pi(f) = [] pi(match) = [2] pi(X) = [] pi(start) = [1] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: top#_A(x1) = (0,0) mark_A(x1) = (0,0) check_A(x1) = ((0,0),(1,0)) x1 + (6,0) found_A(x1) = (1,0) active_A(x1) = (4,5) proper_A(x1) = (4,5) c_A() = (0,0) ok_A(x1) = (2,3) f_A(x1) = (3,4) match_A(x1,x2) = (5,6) X_A() = (4,6) start_A(x1) = ((1,0),(1,1)) x1 precedence: start > c > proper = match > ok > active > f > X > found > top# = mark = check partial status: pi(top#) = [] pi(mark) = [] pi(check) = [] pi(found) = [] pi(active) = [] pi(proper) = [] pi(c) = [] pi(ok) = [] pi(f) = [] pi(match) = [] pi(X) = [] pi(start) = [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: top#(found(x)) -> top#(active(x)) and R consists of: r1: active(f(x)) -> mark(x) r2: top(active(c())) -> top(mark(c())) r3: top(mark(x)) -> top(check(x)) r4: check(f(x)) -> f(check(x)) r5: check(x) -> start(match(f(X()),x)) r6: match(f(x),f(y)) -> f(match(x,y)) r7: match(X(),x) -> proper(x) r8: proper(c()) -> ok(c()) r9: proper(f(x)) -> f(proper(x)) r10: f(ok(x)) -> ok(f(x)) r11: start(ok(x)) -> found(x) r12: f(found(x)) -> found(f(x)) r13: top(found(x)) -> top(active(x)) r14: active(f(x)) -> f(active(x)) r15: f(mark(x)) -> mark(f(x)) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: top#(found(x)) -> top#(active(x)) and R consists of: r1: active(f(x)) -> mark(x) r2: top(active(c())) -> top(mark(c())) r3: top(mark(x)) -> top(check(x)) r4: check(f(x)) -> f(check(x)) r5: check(x) -> start(match(f(X()),x)) r6: match(f(x),f(y)) -> f(match(x,y)) r7: match(X(),x) -> proper(x) r8: proper(c()) -> ok(c()) r9: proper(f(x)) -> f(proper(x)) r10: f(ok(x)) -> ok(f(x)) r11: start(ok(x)) -> found(x) r12: f(found(x)) -> found(f(x)) r13: top(found(x)) -> top(active(x)) r14: active(f(x)) -> f(active(x)) r15: f(mark(x)) -> mark(f(x)) The set of usable rules consists of r1, r10, r12, r14, r15 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: top#_A(x1) = x1 + (1,1) found_A(x1) = ((1,0),(1,1)) x1 + (4,3) active_A(x1) = ((1,0),(1,1)) x1 + (2,1) f_A(x1) = ((1,0),(1,1)) x1 + (3,2) ok_A(x1) = x1 + (4,1) mark_A(x1) = (4,3) precedence: top# > found = active = f = mark > ok partial status: pi(top#) = [1] pi(found) = [1] pi(active) = [1] pi(f) = [1] pi(ok) = [1] pi(mark) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: top#_A(x1) = (0,0) found_A(x1) = ((1,0),(1,1)) x1 + (3,3) active_A(x1) = ((1,0),(1,1)) x1 + (1,1) f_A(x1) = ((1,0),(1,1)) x1 + (2,2) ok_A(x1) = x1 + (1,1) mark_A(x1) = (0,0) precedence: top# > found = active = f = ok = mark partial status: pi(top#) = [] pi(found) = [1] pi(active) = [1] pi(f) = [] pi(ok) = [] pi(mark) = [] 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: check#(f(x)) -> check#(x) and R consists of: r1: active(f(x)) -> mark(x) r2: top(active(c())) -> top(mark(c())) r3: top(mark(x)) -> top(check(x)) r4: check(f(x)) -> f(check(x)) r5: check(x) -> start(match(f(X()),x)) r6: match(f(x),f(y)) -> f(match(x,y)) r7: match(X(),x) -> proper(x) r8: proper(c()) -> ok(c()) r9: proper(f(x)) -> f(proper(x)) r10: f(ok(x)) -> ok(f(x)) r11: start(ok(x)) -> found(x) r12: f(found(x)) -> found(f(x)) r13: top(found(x)) -> top(active(x)) r14: active(f(x)) -> f(active(x)) r15: f(mark(x)) -> mark(f(x)) The set of usable rules consists of (no rules) Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: check#_A(x1) = ((1,0),(1,1)) x1 + (2,2) f_A(x1) = ((1,0),(0,0)) x1 + (1,1) precedence: check# = f partial status: pi(check#) = [] pi(f) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: check#_A(x1) = ((1,0),(0,0)) x1 f_A(x1) = (1,1) precedence: f > check# partial status: pi(check#) = [] pi(f) = [] 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: active#(f(x)) -> active#(x) and R consists of: r1: active(f(x)) -> mark(x) r2: top(active(c())) -> top(mark(c())) r3: top(mark(x)) -> top(check(x)) r4: check(f(x)) -> f(check(x)) r5: check(x) -> start(match(f(X()),x)) r6: match(f(x),f(y)) -> f(match(x,y)) r7: match(X(),x) -> proper(x) r8: proper(c()) -> ok(c()) r9: proper(f(x)) -> f(proper(x)) r10: f(ok(x)) -> ok(f(x)) r11: start(ok(x)) -> found(x) r12: f(found(x)) -> found(f(x)) r13: top(found(x)) -> top(active(x)) r14: active(f(x)) -> f(active(x)) r15: f(mark(x)) -> mark(f(x)) The set of usable rules consists of (no rules) Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: active#_A(x1) = ((1,0),(1,1)) x1 + (2,2) f_A(x1) = ((1,0),(0,0)) x1 + (1,1) precedence: active# = f partial status: pi(active#) = [] pi(f) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: active#_A(x1) = ((1,0),(0,0)) x1 f_A(x1) = (1,1) precedence: f > active# partial status: pi(active#) = [] pi(f) = [] 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: match#(f(x),f(y)) -> match#(x,y) and R consists of: r1: active(f(x)) -> mark(x) r2: top(active(c())) -> top(mark(c())) r3: top(mark(x)) -> top(check(x)) r4: check(f(x)) -> f(check(x)) r5: check(x) -> start(match(f(X()),x)) r6: match(f(x),f(y)) -> f(match(x,y)) r7: match(X(),x) -> proper(x) r8: proper(c()) -> ok(c()) r9: proper(f(x)) -> f(proper(x)) r10: f(ok(x)) -> ok(f(x)) r11: start(ok(x)) -> found(x) r12: f(found(x)) -> found(f(x)) r13: top(found(x)) -> top(active(x)) r14: active(f(x)) -> f(active(x)) r15: f(mark(x)) -> mark(f(x)) The set of usable rules consists of (no rules) Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: match#_A(x1,x2) = ((1,0),(0,0)) x1 f_A(x1) = ((1,0),(0,0)) x1 + (1,1) precedence: match# > f partial status: pi(match#) = [] pi(f) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: match#_A(x1,x2) = (0,0) f_A(x1) = (1,1) precedence: match# = f partial status: pi(match#) = [] pi(f) = [] 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: proper#(f(x)) -> proper#(x) and R consists of: r1: active(f(x)) -> mark(x) r2: top(active(c())) -> top(mark(c())) r3: top(mark(x)) -> top(check(x)) r4: check(f(x)) -> f(check(x)) r5: check(x) -> start(match(f(X()),x)) r6: match(f(x),f(y)) -> f(match(x,y)) r7: match(X(),x) -> proper(x) r8: proper(c()) -> ok(c()) r9: proper(f(x)) -> f(proper(x)) r10: f(ok(x)) -> ok(f(x)) r11: start(ok(x)) -> found(x) r12: f(found(x)) -> found(f(x)) r13: top(found(x)) -> top(active(x)) r14: active(f(x)) -> f(active(x)) r15: f(mark(x)) -> mark(f(x)) The set of usable rules consists of (no rules) Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: proper#_A(x1) = ((1,0),(1,1)) x1 + (2,2) f_A(x1) = ((1,0),(0,0)) x1 + (1,1) precedence: proper# = f partial status: pi(proper#) = [] pi(f) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: proper#_A(x1) = ((1,0),(0,0)) x1 f_A(x1) = (1,1) precedence: f > proper# partial status: pi(proper#) = [] pi(f) = [] 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: f#(ok(x)) -> f#(x) p2: f#(mark(x)) -> f#(x) p3: f#(found(x)) -> f#(x) and R consists of: r1: active(f(x)) -> mark(x) r2: top(active(c())) -> top(mark(c())) r3: top(mark(x)) -> top(check(x)) r4: check(f(x)) -> f(check(x)) r5: check(x) -> start(match(f(X()),x)) r6: match(f(x),f(y)) -> f(match(x,y)) r7: match(X(),x) -> proper(x) r8: proper(c()) -> ok(c()) r9: proper(f(x)) -> f(proper(x)) r10: f(ok(x)) -> ok(f(x)) r11: start(ok(x)) -> found(x) r12: f(found(x)) -> found(f(x)) r13: top(found(x)) -> top(active(x)) r14: active(f(x)) -> f(active(x)) r15: f(mark(x)) -> mark(f(x)) The set of usable rules consists of (no rules) Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: f#_A(x1) = ((1,0),(0,0)) x1 ok_A(x1) = ((1,0),(0,0)) x1 + (1,1) mark_A(x1) = x1 + (1,1) found_A(x1) = ((1,0),(0,0)) x1 + (1,1) precedence: ok > mark = found > f# partial status: pi(f#) = [] pi(ok) = [] pi(mark) = [1] pi(found) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: f#_A(x1) = (0,2) ok_A(x1) = (1,1) mark_A(x1) = ((1,0),(0,0)) x1 + (1,1) found_A(x1) = (1,1) precedence: f# = ok = mark = found partial status: pi(f#) = [] pi(ok) = [] pi(mark) = [] pi(found) = [] The next rules are strictly ordered: p1, p2, p3 We remove them from the problem. Then no dependency pair remains.