YES We show the termination of the TRS R: f(s(x)) -> f(id_inc(c(x,x))) f(c(s(x),y)) -> g(c(x,y)) g(c(s(x),y)) -> g(c(y,x)) g(c(x,s(y))) -> g(c(y,x)) g(c(x,x)) -> f(x) id_inc(c(x,y)) -> c(id_inc(x),id_inc(y)) id_inc(s(x)) -> s(id_inc(x)) id_inc(|0|()) -> |0|() id_inc(|0|()) -> s(|0|()) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: f#(s(x)) -> f#(id_inc(c(x,x))) p2: f#(s(x)) -> id_inc#(c(x,x)) p3: f#(c(s(x),y)) -> g#(c(x,y)) p4: g#(c(s(x),y)) -> g#(c(y,x)) p5: g#(c(x,s(y))) -> g#(c(y,x)) p6: g#(c(x,x)) -> f#(x) p7: id_inc#(c(x,y)) -> id_inc#(x) p8: id_inc#(c(x,y)) -> id_inc#(y) p9: id_inc#(s(x)) -> id_inc#(x) and R consists of: r1: f(s(x)) -> f(id_inc(c(x,x))) r2: f(c(s(x),y)) -> g(c(x,y)) r3: g(c(s(x),y)) -> g(c(y,x)) r4: g(c(x,s(y))) -> g(c(y,x)) r5: g(c(x,x)) -> f(x) r6: id_inc(c(x,y)) -> c(id_inc(x),id_inc(y)) r7: id_inc(s(x)) -> s(id_inc(x)) r8: id_inc(|0|()) -> |0|() r9: id_inc(|0|()) -> s(|0|()) The estimated dependency graph contains the following SCCs: {p1, p3, p4, p5, p6} {p7, p8, p9} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: f#(s(x)) -> f#(id_inc(c(x,x))) p2: f#(c(s(x),y)) -> g#(c(x,y)) p3: g#(c(x,x)) -> f#(x) p4: g#(c(x,s(y))) -> g#(c(y,x)) p5: g#(c(s(x),y)) -> g#(c(y,x)) and R consists of: r1: f(s(x)) -> f(id_inc(c(x,x))) r2: f(c(s(x),y)) -> g(c(x,y)) r3: g(c(s(x),y)) -> g(c(y,x)) r4: g(c(x,s(y))) -> g(c(y,x)) r5: g(c(x,x)) -> f(x) r6: id_inc(c(x,y)) -> c(id_inc(x),id_inc(y)) r7: id_inc(s(x)) -> s(id_inc(x)) r8: id_inc(|0|()) -> |0|() r9: id_inc(|0|()) -> s(|0|()) The set of usable rules consists of r6, r7, r8, r9 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: f#_A(x1) = 1 s_A(x1) = max{26, x1 + 6} id_inc_A(x1) = max{26, x1 + 6} c_A(x1,x2) = max{8, x1 - 3, x2} g#_A(x1) = 1 |0|_A = 27 f#!_A(x1) = max{24, x1 + 17} s!_A(x1) = 75 id_inc!_A(x1) = max{75, x1 + 47} c!_A(x1,x2) = max{24, x1 + 17, x2 + 12} g#!_A(x1) = max{19, x1 + 17} |0|!_A = 76 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: f#_A(x1) = 1 s_A(x1) = max{77, x1 + 37} id_inc_A(x1) = x1 + 39 c_A(x1,x2) = max{23, x1 - 8} g#_A(x1) = 1 |0|_A = 38 f#!_A(x1) = max{56, x1 - 7} s!_A(x1) = 0 id_inc!_A(x1) = 0 c!_A(x1,x2) = x2 + 57 g#!_A(x1) = x1 + 29 |0|!_A = 0 partial status: pi(f#) = [] pi(s) = [] pi(id_inc) = [1] pi(c) = [] pi(g#) = [1] pi(|0|) = [] The following rules are removed: p1, p2, p3, p4, p5 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: id_inc#(c(x,y)) -> id_inc#(x) p2: id_inc#(s(x)) -> id_inc#(x) p3: id_inc#(c(x,y)) -> id_inc#(y) and R consists of: r1: f(s(x)) -> f(id_inc(c(x,x))) r2: f(c(s(x),y)) -> g(c(x,y)) r3: g(c(s(x),y)) -> g(c(y,x)) r4: g(c(x,s(y))) -> g(c(y,x)) r5: g(c(x,x)) -> f(x) r6: id_inc(c(x,y)) -> c(id_inc(x),id_inc(y)) r7: id_inc(s(x)) -> s(id_inc(x)) r8: id_inc(|0|()) -> |0|() r9: id_inc(|0|()) -> s(|0|()) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: id_inc#_A(x1) = 0 c_A(x1,x2) = max{1, x1 - 5, x2 - 5} s_A(x1) = max{7, x1 - 5} id_inc#!_A(x1) = 0 c!_A(x1,x2) = max{1, x1 - 14} s!_A(x1) = max{2, x1 - 12} 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: id_inc#_A(x1) = 0 c_A(x1,x2) = max{0, x1 - 15, x2 - 9} s_A(x1) = max{21, x1 + 5} id_inc#!_A(x1) = max{3, x1 - 1} c!_A(x1,x2) = max{1, x1 - 23, x2 - 30} s!_A(x1) = max{2, x1 - 7} partial status: pi(id_inc#) = [1] pi(c) = [1, 2] pi(s) = [1] The following rules are removed: p1, p2, p3 Then no dependency pair remains.