YES We show the termination of the TRS R: app(app(plus(),|0|()),y) -> y app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y)) app(app(times(),|0|()),y) -> |0|() app(app(times(),app(s(),x)),y) -> app(app(plus(),app(app(times(),x),y)),y) app(inc(),xs) -> app(app(map(),app(plus(),app(s(),|0|()))),xs) app(double(),xs) -> app(app(map(),app(times(),app(s(),app(s(),|0|())))),xs) app(app(map(),f),nil()) -> nil() app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs)) Instead we prove termination of the uncurried system: app(cons_0(),y) -> cons_1(y) app(cons_1(x1),y) -> cons_2(x1,y) app(double_0(),y) -> double_1(y) app(inc_0(),y) -> inc_1(y) app(map_0(),y) -> map_1(y) app(map_1(x1),y) -> map_2(x1,y) app(plus_0(),y) -> plus_1(y) app(plus_1(x1),y) -> plus_2(x1,y) app(s_0(),y) -> s_1(y) app(times_0(),y) -> times_1(y) app(times_1(x1),y) -> times_2(x1,y) plus_2(|0|_0(),y) -> y plus_2(s_1(x),y) -> s_1(plus_2(x,y)) times_2(|0|_0(),y) -> |0|_0() times_2(s_1(x),y) -> plus_2(times_2(x,y),y) inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs) map_2(f,nil_0()) -> nil_0() map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(double_0(),y) -> double_1#(y) p2: app#(inc_0(),y) -> inc_1#(y) p3: app#(map_1(x1),y) -> map_2#(x1,y) p4: app#(plus_1(x1),y) -> plus_2#(x1,y) p5: app#(times_1(x1),y) -> times_2#(x1,y) p6: plus_2#(s_1(x),y) -> plus_2#(x,y) p7: times_2#(s_1(x),y) -> plus_2#(times_2(x,y),y) p8: times_2#(s_1(x),y) -> times_2#(x,y) p9: inc_1#(xs) -> map_2#(plus_1(s_1(|0|_0())),xs) p10: double_1#(xs) -> map_2#(times_1(s_1(s_1(|0|_0()))),xs) p11: map_2#(f,cons_2(x,xs)) -> app#(f,x) p12: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(double_0(),y) -> double_1(y) r4: app(inc_0(),y) -> inc_1(y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(plus_0(),y) -> plus_1(y) r8: app(plus_1(x1),y) -> plus_2(x1,y) r9: app(s_0(),y) -> s_1(y) r10: app(times_0(),y) -> times_1(y) r11: app(times_1(x1),y) -> times_2(x1,y) r12: plus_2(|0|_0(),y) -> y r13: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r14: times_2(|0|_0(),y) -> |0|_0() r15: times_2(s_1(x),y) -> plus_2(times_2(x,y),y) r16: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r17: double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs) r18: map_2(f,nil_0()) -> nil_0() r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p9, p10, p11, p12} {p8} {p6} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(double_0(),y) -> double_1#(y) p2: double_1#(xs) -> map_2#(times_1(s_1(s_1(|0|_0()))),xs) p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p4: map_2#(f,cons_2(x,xs)) -> app#(f,x) p5: app#(map_1(x1),y) -> map_2#(x1,y) p6: app#(inc_0(),y) -> inc_1#(y) p7: inc_1#(xs) -> map_2#(plus_1(s_1(|0|_0())),xs) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(double_0(),y) -> double_1(y) r4: app(inc_0(),y) -> inc_1(y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(plus_0(),y) -> plus_1(y) r8: app(plus_1(x1),y) -> plus_2(x1,y) r9: app(s_0(),y) -> s_1(y) r10: app(times_0(),y) -> times_1(y) r11: app(times_1(x1),y) -> times_2(x1,y) r12: plus_2(|0|_0(),y) -> y r13: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r14: times_2(|0|_0(),y) -> |0|_0() r15: times_2(s_1(x),y) -> plus_2(times_2(x,y),y) r16: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r17: double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs) r18: map_2(f,nil_0()) -> nil_0() r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The set of usable rules consists of r1, r5, r7, r9, r10 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: app#_A(x1,x2) = x1 + 3 double_0_A() = 11 double_1#_A(x1) = 8 map_2#_A(x1,x2) = x1 + 4 times_1_A(x1) = 3 s_1_A(x1) = 9 |0|_0_A() = 1 cons_2_A(x1,x2) = x1 + x2 + 2 map_1_A(x1) = x1 + 5 inc_0_A() = 11 inc_1#_A(x1) = 8 plus_1_A(x1) = 3 app_A(x1,x2) = x2 + 11 cons_0_A() = 2 cons_1_A(x1) = 1 map_0_A() = 1 plus_0_A() = 4 s_0_A() = 10 times_0_A() = 4 base order 2: precedence: app# = inc_0 > inc_1# = cons_0 = plus_0 > plus_1 = cons_1 > map_2# = app > double_0 = map_1 > cons_2 = times_0 > times_1 = s_1 = map_0 > s_0 > |0|_0 > double_1# partial status: pi(app#) = [1] pi(double_0) = [] pi(double_1#) = [] pi(map_2#) = [] pi(times_1) = [] pi(s_1) = [] pi(|0|_0) = [] pi(cons_2) = [2] pi(map_1) = [1] pi(inc_0) = [] pi(inc_1#) = [] pi(plus_1) = [] pi(app) = [2] pi(cons_0) = [] pi(cons_1) = [] pi(map_0) = [] pi(plus_0) = [] pi(s_0) = [] pi(times_0) = [] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: double_1#(xs) -> map_2#(times_1(s_1(s_1(|0|_0()))),xs) p2: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p3: map_2#(f,cons_2(x,xs)) -> app#(f,x) p4: app#(map_1(x1),y) -> map_2#(x1,y) p5: app#(inc_0(),y) -> inc_1#(y) p6: inc_1#(xs) -> map_2#(plus_1(s_1(|0|_0())),xs) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(double_0(),y) -> double_1(y) r4: app(inc_0(),y) -> inc_1(y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(plus_0(),y) -> plus_1(y) r8: app(plus_1(x1),y) -> plus_2(x1,y) r9: app(s_0(),y) -> s_1(y) r10: app(times_0(),y) -> times_1(y) r11: app(times_1(x1),y) -> times_2(x1,y) r12: plus_2(|0|_0(),y) -> y r13: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r14: times_2(|0|_0(),y) -> |0|_0() r15: times_2(s_1(x),y) -> plus_2(times_2(x,y),y) r16: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r17: double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs) r18: map_2(f,nil_0()) -> nil_0() r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The estimated dependency graph contains the following SCCs: {p2, p3, p4, p5, p6} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p2: map_2#(f,cons_2(x,xs)) -> app#(f,x) p3: app#(inc_0(),y) -> inc_1#(y) p4: inc_1#(xs) -> map_2#(plus_1(s_1(|0|_0())),xs) p5: app#(map_1(x1),y) -> map_2#(x1,y) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(double_0(),y) -> double_1(y) r4: app(inc_0(),y) -> inc_1(y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(plus_0(),y) -> plus_1(y) r8: app(plus_1(x1),y) -> plus_2(x1,y) r9: app(s_0(),y) -> s_1(y) r10: app(times_0(),y) -> times_1(y) r11: app(times_1(x1),y) -> times_2(x1,y) r12: plus_2(|0|_0(),y) -> y r13: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r14: times_2(|0|_0(),y) -> |0|_0() r15: times_2(s_1(x),y) -> plus_2(times_2(x,y),y) r16: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r17: double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs) r18: map_2(f,nil_0()) -> nil_0() r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The set of usable rules consists of r1, r5, r7, r9, r10, r13 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: map_2#_A(x1,x2) = x1 + 1 cons_2_A(x1,x2) = x1 + x2 + 2 app#_A(x1,x2) = x1 inc_0_A() = 6 inc_1#_A(x1) = 5 plus_1_A(x1) = 3 s_1_A(x1) = 1 |0|_0_A() = 1 map_1_A(x1) = x1 + 2 app_A(x1,x2) = x2 + 5 cons_0_A() = 2 cons_1_A(x1) = 1 map_0_A() = 1 plus_0_A() = 4 s_0_A() = 2 times_0_A() = 4 times_1_A(x1) = 3 plus_2_A(x1,x2) = 2 base order 2: precedence: inc_0 > inc_1# = plus_0 = times_0 > app# = plus_1 = |0|_0 = times_1 > app = s_0 > s_1 = map_1 = plus_2 > map_2# = map_0 > cons_2 > cons_1 > cons_0 partial status: pi(map_2#) = [] pi(cons_2) = [1, 2] pi(app#) = [1] pi(inc_0) = [] pi(inc_1#) = [] pi(plus_1) = [] pi(s_1) = [] pi(|0|_0) = [] pi(map_1) = [1] pi(app) = [2] pi(cons_0) = [] pi(cons_1) = [] pi(map_0) = [] pi(plus_0) = [] pi(s_0) = [] pi(times_0) = [] pi(times_1) = [] pi(plus_2) = [] The following rules are removed: p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p2: app#(inc_0(),y) -> inc_1#(y) p3: inc_1#(xs) -> map_2#(plus_1(s_1(|0|_0())),xs) p4: app#(map_1(x1),y) -> map_2#(x1,y) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(double_0(),y) -> double_1(y) r4: app(inc_0(),y) -> inc_1(y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(plus_0(),y) -> plus_1(y) r8: app(plus_1(x1),y) -> plus_2(x1,y) r9: app(s_0(),y) -> s_1(y) r10: app(times_0(),y) -> times_1(y) r11: app(times_1(x1),y) -> times_2(x1,y) r12: plus_2(|0|_0(),y) -> y r13: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r14: times_2(|0|_0(),y) -> |0|_0() r15: times_2(s_1(x),y) -> plus_2(times_2(x,y),y) r16: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r17: double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs) r18: map_2(f,nil_0()) -> nil_0() r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(double_0(),y) -> double_1(y) r4: app(inc_0(),y) -> inc_1(y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(plus_0(),y) -> plus_1(y) r8: app(plus_1(x1),y) -> plus_2(x1,y) r9: app(s_0(),y) -> s_1(y) r10: app(times_0(),y) -> times_1(y) r11: app(times_1(x1),y) -> times_2(x1,y) r12: plus_2(|0|_0(),y) -> y r13: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r14: times_2(|0|_0(),y) -> |0|_0() r15: times_2(s_1(x),y) -> plus_2(times_2(x,y),y) r16: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r17: double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs) r18: map_2(f,nil_0()) -> nil_0() r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The set of usable rules consists of r1, r7, r9, r10 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: map_2#_A(x1,x2) = x2 cons_2_A(x1,x2) = x1 + x2 + 6 app_A(x1,x2) = x2 + 7 cons_0_A() = 2 cons_1_A(x1) = 1 plus_0_A() = 2 plus_1_A(x1) = 1 s_0_A() = 4 s_1_A(x1) = 3 times_0_A() = 6 times_1_A(x1) = 5 base order 2: precedence: times_0 > times_1 > cons_2 = app = cons_0 = plus_0 = s_0 > map_2# = cons_1 > plus_1 = s_1 partial status: pi(map_2#) = [] pi(cons_2) = [2] pi(app) = [2] pi(cons_0) = [] pi(cons_1) = [] pi(plus_0) = [] pi(plus_1) = [] pi(s_0) = [] pi(s_1) = [] pi(times_0) = [] pi(times_1) = [] 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: times_2#(s_1(x),y) -> times_2#(x,y) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(double_0(),y) -> double_1(y) r4: app(inc_0(),y) -> inc_1(y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(plus_0(),y) -> plus_1(y) r8: app(plus_1(x1),y) -> plus_2(x1,y) r9: app(s_0(),y) -> s_1(y) r10: app(times_0(),y) -> times_1(y) r11: app(times_1(x1),y) -> times_2(x1,y) r12: plus_2(|0|_0(),y) -> y r13: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r14: times_2(|0|_0(),y) -> |0|_0() r15: times_2(s_1(x),y) -> plus_2(times_2(x,y),y) r16: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r17: double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs) r18: map_2(f,nil_0()) -> nil_0() r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The set of usable rules consists of r1, r5, r7, r9, r10, r12, r13 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: times_2#_A(x1,x2) = x1 + x2 + 3 s_1_A(x1) = x1 + 2 app_A(x1,x2) = x2 + 7 cons_0_A() = 2 cons_1_A(x1) = 1 map_0_A() = 2 map_1_A(x1) = 1 plus_0_A() = 2 plus_1_A(x1) = 1 s_0_A() = 1 times_0_A() = 2 times_1_A(x1) = 1 plus_2_A(x1,x2) = x1 + x2 + 8 |0|_0_A() = 1 base order 2: precedence: times_2# = plus_2 > s_1 = map_0 = plus_0 = times_0 > app = cons_1 = s_0 > cons_0 = map_1 > plus_1 = times_1 = |0|_0 partial status: pi(times_2#) = [1, 2] pi(s_1) = [] pi(app) = [2] pi(cons_0) = [] pi(cons_1) = [] pi(map_0) = [] pi(map_1) = [] pi(plus_0) = [] pi(plus_1) = [] pi(s_0) = [] pi(times_0) = [] pi(times_1) = [] pi(plus_2) = [1, 2] pi(|0|_0) = [] 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: plus_2#(s_1(x),y) -> plus_2#(x,y) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(double_0(),y) -> double_1(y) r4: app(inc_0(),y) -> inc_1(y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(plus_0(),y) -> plus_1(y) r8: app(plus_1(x1),y) -> plus_2(x1,y) r9: app(s_0(),y) -> s_1(y) r10: app(times_0(),y) -> times_1(y) r11: app(times_1(x1),y) -> times_2(x1,y) r12: plus_2(|0|_0(),y) -> y r13: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r14: times_2(|0|_0(),y) -> |0|_0() r15: times_2(s_1(x),y) -> plus_2(times_2(x,y),y) r16: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r17: double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs) r18: map_2(f,nil_0()) -> nil_0() r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The set of usable rules consists of r1, r5, r7, r9, r10, r12, r13 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: plus_2#_A(x1,x2) = x1 + x2 + 3 s_1_A(x1) = x1 + 2 app_A(x1,x2) = x2 + 7 cons_0_A() = 2 cons_1_A(x1) = 1 map_0_A() = 2 map_1_A(x1) = 1 plus_0_A() = 2 plus_1_A(x1) = 1 s_0_A() = 1 times_0_A() = 2 times_1_A(x1) = 1 plus_2_A(x1,x2) = x1 + x2 + 8 |0|_0_A() = 1 base order 2: precedence: plus_2# = plus_2 > s_1 = map_0 = plus_0 = times_0 > app = cons_1 = s_0 > cons_0 = map_1 > plus_1 = times_1 = |0|_0 partial status: pi(plus_2#) = [1, 2] pi(s_1) = [] pi(app) = [2] pi(cons_0) = [] pi(cons_1) = [] pi(map_0) = [] pi(map_1) = [] pi(plus_0) = [] pi(plus_1) = [] pi(s_0) = [] pi(times_0) = [] pi(times_1) = [] pi(plus_2) = [1, 2] pi(|0|_0) = [] The following rules are removed: p1 Then no dependency pair remains.