YES We show the termination of the TRS R: f(|0|()) -> true() f(|1|()) -> false() f(s(x)) -> f(x) if(true(),s(x),s(y)) -> s(x) if(false(),s(x),s(y)) -> s(y) g(x,c(y)) -> c(g(x,y)) g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y))) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: f#(s(x)) -> f#(x) p2: g#(x,c(y)) -> g#(x,y) p3: g#(x,c(y)) -> g#(x,if(f(x),c(g(s(x),y)),c(y))) p4: g#(x,c(y)) -> if#(f(x),c(g(s(x),y)),c(y)) p5: g#(x,c(y)) -> f#(x) p6: g#(x,c(y)) -> g#(s(x),y) and R consists of: r1: f(|0|()) -> true() r2: f(|1|()) -> false() r3: f(s(x)) -> f(x) r4: if(true(),s(x),s(y)) -> s(x) r5: if(false(),s(x),s(y)) -> s(y) r6: g(x,c(y)) -> c(g(x,y)) r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y))) The estimated dependency graph contains the following SCCs: {p2, p6} {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: g#(x,c(y)) -> g#(s(x),y) p2: g#(x,c(y)) -> g#(x,y) and R consists of: r1: f(|0|()) -> true() r2: f(|1|()) -> false() r3: f(s(x)) -> f(x) r4: if(true(),s(x),s(y)) -> s(x) r5: if(false(),s(x),s(y)) -> s(y) r6: g(x,c(y)) -> c(g(x,y)) r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y))) The set of usable rules consists of r1, r2, r3, r5 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: g#_A(x1,x2) = x1 + x2 + 6 c_A(x1) = x1 + 5 s_A(x1) = x1 + 4 f_A(x1) = 2 |0|_A() = 2 true_A() = 1 |1|_A() = 2 false_A() = 1 if_A(x1,x2,x3) = x3 base order 2: precedence: g# > c > if > s > f = |0| = |1| > true = false partial status: pi(g#) = [] pi(c) = [1] pi(s) = [1] pi(f) = [] pi(|0|) = [] pi(true) = [] pi(|1|) = [] pi(false) = [] pi(if) = [] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: g#(x,c(y)) -> g#(x,y) and R consists of: r1: f(|0|()) -> true() r2: f(|1|()) -> false() r3: f(s(x)) -> f(x) r4: if(true(),s(x),s(y)) -> s(x) r5: if(false(),s(x),s(y)) -> s(y) r6: g(x,c(y)) -> c(g(x,y)) r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y))) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: g#(x,c(y)) -> g#(x,y) and R consists of: r1: f(|0|()) -> true() r2: f(|1|()) -> false() r3: f(s(x)) -> f(x) r4: if(true(),s(x),s(y)) -> s(x) r5: if(false(),s(x),s(y)) -> s(y) r6: g(x,c(y)) -> c(g(x,y)) r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y))) The set of usable rules consists of r1, r2, r3, r4, r5 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: g#_A(x1,x2) = x1 + x2 + 4 c_A(x1) = x1 + 3 f_A(x1) = 2 |0|_A() = 2 true_A() = 1 |1|_A() = 2 false_A() = 1 s_A(x1) = 3 if_A(x1,x2,x3) = x3 base order 2: precedence: g# > c > if > s > f = |0| = |1| > true = false partial status: pi(g#) = [1] pi(c) = [1] pi(f) = [] pi(|0|) = [] pi(true) = [] pi(|1|) = [] pi(false) = [] pi(s) = [] pi(if) = [] The following rules are removed: p1 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: f#(s(x)) -> f#(x) and R consists of: r1: f(|0|()) -> true() r2: f(|1|()) -> false() r3: f(s(x)) -> f(x) r4: if(true(),s(x),s(y)) -> s(x) r5: if(false(),s(x),s(y)) -> s(y) r6: g(x,c(y)) -> c(g(x,y)) r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y))) The set of usable rules consists of r1, r2, r3 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: f#_A(x1) = x1 s_A(x1) = x1 + 5 f_A(x1) = 2 |0|_A() = 2 true_A() = 1 |1|_A() = 2 false_A() = 1 base order 2: precedence: f# > s > |0| > f = |1| > true = false partial status: pi(f#) = [1] pi(s) = [1] pi(f) = [] pi(|0|) = [] pi(true) = [] pi(|1|) = [] pi(false) = [] The following rules are removed: p1 Then no dependency pair remains.