YES We show the termination of the TRS R: nonZero(|0|()) -> false() nonZero(s(x)) -> true() p(|0|()) -> |0|() p(s(x)) -> x id_inc(x) -> x id_inc(x) -> s(x) random(x) -> rand(x,|0|()) rand(x,y) -> if(nonZero(x),x,y) if(false(),x,y) -> y if(true(),x,y) -> rand(p(x),id_inc(y)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: random#(x) -> rand#(x,|0|()) p2: rand#(x,y) -> if#(nonZero(x),x,y) p3: rand#(x,y) -> nonZero#(x) p4: if#(true(),x,y) -> rand#(p(x),id_inc(y)) p5: if#(true(),x,y) -> p#(x) p6: if#(true(),x,y) -> id_inc#(y) and R consists of: r1: nonZero(|0|()) -> false() r2: nonZero(s(x)) -> true() r3: p(|0|()) -> |0|() r4: p(s(x)) -> x r5: id_inc(x) -> x r6: id_inc(x) -> s(x) r7: random(x) -> rand(x,|0|()) r8: rand(x,y) -> if(nonZero(x),x,y) r9: if(false(),x,y) -> y r10: if(true(),x,y) -> rand(p(x),id_inc(y)) The estimated dependency graph contains the following SCCs: {p2, p4} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: rand#(x,y) -> if#(nonZero(x),x,y) p2: if#(true(),x,y) -> rand#(p(x),id_inc(y)) and R consists of: r1: nonZero(|0|()) -> false() r2: nonZero(s(x)) -> true() r3: p(|0|()) -> |0|() r4: p(s(x)) -> x r5: id_inc(x) -> x r6: id_inc(x) -> s(x) r7: random(x) -> rand(x,|0|()) r8: rand(x,y) -> if(nonZero(x),x,y) r9: if(false(),x,y) -> y r10: if(true(),x,y) -> rand(p(x),id_inc(y)) The set of usable rules consists of r1, r2, r3, r4, r5, r6 Take the reduction pair: max/plus interpretations on natural numbers: rand#_A(x1,x2) = x1 + 4 if#_A(x1,x2,x3) = max{x1 + 2, x2 + 3} nonZero_A(x1) = max{1, x1} true_A = 2 p_A(x1) = max{0, x1 - 1} id_inc_A(x1) = max{4, x1 + 2} |0|_A = 0 false_A = 1 s_A(x1) = max{3, x1 + 2} 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: if#(true(),x,y) -> rand#(p(x),id_inc(y)) and R consists of: r1: nonZero(|0|()) -> false() r2: nonZero(s(x)) -> true() r3: p(|0|()) -> |0|() r4: p(s(x)) -> x r5: id_inc(x) -> x r6: id_inc(x) -> s(x) r7: random(x) -> rand(x,|0|()) r8: rand(x,y) -> if(nonZero(x),x,y) r9: if(false(),x,y) -> y r10: if(true(),x,y) -> rand(p(x),id_inc(y)) The estimated dependency graph contains the following SCCs: (no SCCs)