YES We show the termination of the TRS R: cond1(true(),x,y) -> cond2(gr(y,|0|()),x,y) cond2(true(),x,y) -> cond2(gr(y,|0|()),x,p(y)) cond2(false(),x,y) -> cond1(gr(x,|0|()),p(x),y) gr(|0|(),x) -> false() gr(s(x),|0|()) -> true() gr(s(x),s(y)) -> gr(x,y) p(|0|()) -> |0|() p(s(x)) -> x -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: cond1#(true(),x,y) -> cond2#(gr(y,|0|()),x,y) p2: cond1#(true(),x,y) -> gr#(y,|0|()) p3: cond2#(true(),x,y) -> cond2#(gr(y,|0|()),x,p(y)) p4: cond2#(true(),x,y) -> gr#(y,|0|()) p5: cond2#(true(),x,y) -> p#(y) p6: cond2#(false(),x,y) -> cond1#(gr(x,|0|()),p(x),y) p7: cond2#(false(),x,y) -> gr#(x,|0|()) p8: cond2#(false(),x,y) -> p#(x) p9: gr#(s(x),s(y)) -> gr#(x,y) and R consists of: r1: cond1(true(),x,y) -> cond2(gr(y,|0|()),x,y) r2: cond2(true(),x,y) -> cond2(gr(y,|0|()),x,p(y)) r3: cond2(false(),x,y) -> cond1(gr(x,|0|()),p(x),y) r4: gr(|0|(),x) -> false() r5: gr(s(x),|0|()) -> true() r6: gr(s(x),s(y)) -> gr(x,y) r7: p(|0|()) -> |0|() r8: p(s(x)) -> x The estimated dependency graph contains the following SCCs: {p1, p3, p6} {p9} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: cond1#(true(),x,y) -> cond2#(gr(y,|0|()),x,y) p2: cond2#(false(),x,y) -> cond1#(gr(x,|0|()),p(x),y) p3: cond2#(true(),x,y) -> cond2#(gr(y,|0|()),x,p(y)) and R consists of: r1: cond1(true(),x,y) -> cond2(gr(y,|0|()),x,y) r2: cond2(true(),x,y) -> cond2(gr(y,|0|()),x,p(y)) r3: cond2(false(),x,y) -> cond1(gr(x,|0|()),p(x),y) r4: gr(|0|(),x) -> false() r5: gr(s(x),|0|()) -> true() r6: gr(s(x),s(y)) -> gr(x,y) r7: p(|0|()) -> |0|() r8: p(s(x)) -> x The set of usable rules consists of r4, r5, r6, r7, r8 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: cond1#_A(x1,x2,x3) = max{x1 - 18, x2 + 4} true_A = 28 cond2#_A(x1,x2,x3) = max{9, x2 + 3} gr_A(x1,x2) = max{17, x1 - 2, x2 + 10} |0|_A = 1 false_A = 16 p_A(x1) = max{4, x1 - 2} s_A(x1) = x1 + 31 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: cond1#_A(x1,x2,x3) = 0 true_A() = 3 cond2#_A(x1,x2,x3) = 0 gr_A(x1,x2) = x2 + 1 |0|_A() = 2 false_A() = 1 p_A(x1) = x1 + 1 s_A(x1) = x1 + 3 cond1#!_A(x1,x2,x3) = x3 + 3 true!_A() = 3 cond2#!_A(x1,x2,x3) = 2 gr!_A(x1,x2) = x1 |0|!_A() = 1 false!_A() = 1 p!_A(x1) = 3 s!_A(x1) = 4 partial status: pi(cond1#) = [] pi(true) = [] pi(cond2#) = [] pi(gr) = [] pi(|0|) = [] pi(false) = [] pi(p) = [] pi(s) = [1] The following rules are removed: p1, p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: cond2#(true(),x,y) -> cond2#(gr(y,|0|()),x,p(y)) and R consists of: r1: cond1(true(),x,y) -> cond2(gr(y,|0|()),x,y) r2: cond2(true(),x,y) -> cond2(gr(y,|0|()),x,p(y)) r3: cond2(false(),x,y) -> cond1(gr(x,|0|()),p(x),y) r4: gr(|0|(),x) -> false() r5: gr(s(x),|0|()) -> true() r6: gr(s(x),s(y)) -> gr(x,y) r7: p(|0|()) -> |0|() r8: p(s(x)) -> x The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: cond2#(true(),x,y) -> cond2#(gr(y,|0|()),x,p(y)) and R consists of: r1: cond1(true(),x,y) -> cond2(gr(y,|0|()),x,y) r2: cond2(true(),x,y) -> cond2(gr(y,|0|()),x,p(y)) r3: cond2(false(),x,y) -> cond1(gr(x,|0|()),p(x),y) r4: gr(|0|(),x) -> false() r5: gr(s(x),|0|()) -> true() r6: gr(s(x),s(y)) -> gr(x,y) r7: p(|0|()) -> |0|() r8: p(s(x)) -> x The set of usable rules consists of r4, r5, r7, r8 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: cond2#_A(x1,x2,x3) = max{15, x1 + 2, x3 - 1} true_A = 14 gr_A(x1,x2) = max{11, x1 - 4} |0|_A = 5 p_A(x1) = max{6, x1 - 2} false_A = 4 s_A(x1) = x1 + 19 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: cond2#_A(x1,x2,x3) = 0 true_A() = 1 gr_A(x1,x2) = x1 + x2 |0|_A() = 1 p_A(x1) = x1 + 2 false_A() = 1 s_A(x1) = x1 + 1 cond2#!_A(x1,x2,x3) = 8 true!_A() = 2 gr!_A(x1,x2) = x2 + 4 |0|!_A() = 1 p!_A(x1) = 7 false!_A() = 0 s!_A(x1) = 3 partial status: pi(cond2#) = [1] pi(true) = [] pi(gr) = [] pi(|0|) = [] pi(p) = [] pi(false) = [] pi(s) = [] 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: gr#(s(x),s(y)) -> gr#(x,y) and R consists of: r1: cond1(true(),x,y) -> cond2(gr(y,|0|()),x,y) r2: cond2(true(),x,y) -> cond2(gr(y,|0|()),x,p(y)) r3: cond2(false(),x,y) -> cond1(gr(x,|0|()),p(x),y) r4: gr(|0|(),x) -> false() r5: gr(s(x),|0|()) -> true() r6: gr(s(x),s(y)) -> gr(x,y) r7: p(|0|()) -> |0|() r8: p(s(x)) -> x The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: gr#_A(x1,x2) = 0 s_A(x1) = max{4, x1 + 3} base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: gr#_A(x1,x2) = x1 + x2 s_A(x1) = x1 + 1 gr#!_A(x1,x2) = x1 s!_A(x1) = 1 partial status: pi(gr#) = [] pi(s) = [1] The following rules are removed: p1 Then no dependency pair remains.