YES We show the termination of the TRS R: not(true()) -> false() not(false()) -> true() evenodd(x,|0|()) -> not(evenodd(x,s(|0|()))) evenodd(|0|(),s(|0|())) -> false() evenodd(s(x),s(|0|())) -> evenodd(x,|0|()) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: evenodd#(x,|0|()) -> not#(evenodd(x,s(|0|()))) p2: evenodd#(x,|0|()) -> evenodd#(x,s(|0|())) p3: evenodd#(s(x),s(|0|())) -> evenodd#(x,|0|()) and R consists of: r1: not(true()) -> false() r2: not(false()) -> true() r3: evenodd(x,|0|()) -> not(evenodd(x,s(|0|()))) r4: evenodd(|0|(),s(|0|())) -> false() r5: evenodd(s(x),s(|0|())) -> evenodd(x,|0|()) The estimated dependency graph contains the following SCCs: {p2, p3} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: evenodd#(x,|0|()) -> evenodd#(x,s(|0|())) p2: evenodd#(s(x),s(|0|())) -> evenodd#(x,|0|()) and R consists of: r1: not(true()) -> false() r2: not(false()) -> true() r3: evenodd(x,|0|()) -> not(evenodd(x,s(|0|()))) r4: evenodd(|0|(),s(|0|())) -> false() r5: evenodd(s(x),s(|0|())) -> evenodd(x,|0|()) The set of usable rules consists of (no rules) Take the reduction pair: weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: evenodd#_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (1,9) |0|_A() = (3,3) s_A(x1) = ((0,0),(1,1)) x1 + (1,2) precedence: |0| > evenodd# = s partial status: pi(evenodd#) = [] pi(|0|) = [] pi(s) = [] 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: evenodd#(s(x),s(|0|())) -> evenodd#(x,|0|()) and R consists of: r1: not(true()) -> false() r2: not(false()) -> true() r3: evenodd(x,|0|()) -> not(evenodd(x,s(|0|()))) r4: evenodd(|0|(),s(|0|())) -> false() r5: evenodd(s(x),s(|0|())) -> evenodd(x,|0|()) The estimated dependency graph contains the following SCCs: (no SCCs)