YES We show the termination of the TRS R: minus(x,|0|()) -> x minus(s(x),s(y)) -> minus(x,y) double(|0|()) -> |0|() double(s(x)) -> s(s(double(x))) plus(|0|(),y) -> y plus(s(x),y) -> s(plus(x,y)) plus(s(x),y) -> plus(x,s(y)) plus(s(x),y) -> s(plus(minus(x,y),double(y))) plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: minus#(s(x),s(y)) -> minus#(x,y) p2: double#(s(x)) -> double#(x) p3: plus#(s(x),y) -> plus#(x,y) p4: plus#(s(x),y) -> plus#(x,s(y)) p5: plus#(s(x),y) -> plus#(minus(x,y),double(y)) p6: plus#(s(x),y) -> minus#(x,y) p7: plus#(s(x),y) -> double#(y) p8: plus#(s(plus(x,y)),z) -> plus#(plus(x,y),z) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(|0|(),y) -> y r6: plus(s(x),y) -> s(plus(x,y)) r7: plus(s(x),y) -> plus(x,s(y)) r8: plus(s(x),y) -> s(plus(minus(x,y),double(y))) r9: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z)) The estimated dependency graph contains the following SCCs: {p3, p4, p5, p8} {p1} {p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: plus#(s(plus(x,y)),z) -> plus#(plus(x,y),z) p2: plus#(s(x),y) -> plus#(minus(x,y),double(y)) p3: plus#(s(x),y) -> plus#(x,s(y)) p4: plus#(s(x),y) -> plus#(x,y) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(|0|(),y) -> y r6: plus(s(x),y) -> s(plus(x,y)) r7: plus(s(x),y) -> plus(x,s(y)) r8: plus(s(x),y) -> s(plus(minus(x,y),double(y))) r9: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: plus#_A(x1,x2) = max{x1 + 11, x2 + 6} s_A(x1) = max{2, x1} plus_A(x1,x2) = max{x1 + 15, x2 + 4} minus_A(x1,x2) = max{1, x1} double_A(x1) = max{7, x1} |0|_A = 1 base order 2: precedence: double > plus > plus# = s = |0| > minus partial status: pi(plus#) = [1] pi(s) = [1] pi(plus) = [1] pi(minus) = [1] pi(double) = [1] pi(|0|) = [] The following rules are removed: p3 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: plus#(s(plus(x,y)),z) -> plus#(plus(x,y),z) p2: plus#(s(x),y) -> plus#(minus(x,y),double(y)) p3: plus#(s(x),y) -> plus#(x,y) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(|0|(),y) -> y r6: plus(s(x),y) -> s(plus(x,y)) r7: plus(s(x),y) -> plus(x,s(y)) r8: plus(s(x),y) -> s(plus(minus(x,y),double(y))) r9: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z)) The estimated dependency graph contains the following SCCs: {p1, p2, p3} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: plus#(s(plus(x,y)),z) -> plus#(plus(x,y),z) p2: plus#(s(x),y) -> plus#(x,y) p3: plus#(s(x),y) -> plus#(minus(x,y),double(y)) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(|0|(),y) -> y r6: plus(s(x),y) -> s(plus(x,y)) r7: plus(s(x),y) -> plus(x,s(y)) r8: plus(s(x),y) -> s(plus(minus(x,y),double(y))) r9: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: plus#_A(x1,x2) = max{2, x1} s_A(x1) = max{2, x1} plus_A(x1,x2) = max{11, x1 + 6, x2 + 8} minus_A(x1,x2) = max{1, x1} double_A(x1) = max{2, x1} |0|_A = 1 base order 2: precedence: plus > double = |0| > s > plus# > minus partial status: pi(plus#) = [1] pi(s) = [1] pi(plus) = [1] pi(minus) = [1] pi(double) = [1] pi(|0|) = [] The following rules are removed: p3 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: plus#(s(plus(x,y)),z) -> plus#(plus(x,y),z) p2: plus#(s(x),y) -> plus#(x,y) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(|0|(),y) -> y r6: plus(s(x),y) -> s(plus(x,y)) r7: plus(s(x),y) -> plus(x,s(y)) r8: plus(s(x),y) -> s(plus(minus(x,y),double(y))) r9: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z)) The estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: plus#(s(plus(x,y)),z) -> plus#(plus(x,y),z) p2: plus#(s(x),y) -> plus#(x,y) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(|0|(),y) -> y r6: plus(s(x),y) -> s(plus(x,y)) r7: plus(s(x),y) -> plus(x,s(y)) r8: plus(s(x),y) -> s(plus(minus(x,y),double(y))) r9: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: plus#_A(x1,x2) = max{1, x1} s_A(x1) = max{14, x1} plus_A(x1,x2) = max{9, x1 + 4, x2 + 2} minus_A(x1,x2) = max{13, x1} |0|_A = 1 double_A(x1) = x1 base order 2: precedence: double > plus > s = |0| > plus# = minus partial status: pi(plus#) = [1] pi(s) = [1] pi(plus) = [1] pi(minus) = [1] pi(|0|) = [] pi(double) = [1] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: plus#(s(x),y) -> plus#(x,y) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(|0|(),y) -> y r6: plus(s(x),y) -> s(plus(x,y)) r7: plus(s(x),y) -> plus(x,s(y)) r8: plus(s(x),y) -> s(plus(minus(x,y),double(y))) r9: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z)) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: plus#(s(x),y) -> plus#(x,y) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(|0|(),y) -> y r6: plus(s(x),y) -> s(plus(x,y)) r7: plus(s(x),y) -> plus(x,s(y)) r8: plus(s(x),y) -> s(plus(minus(x,y),double(y))) r9: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z)) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: plus#_A(x1,x2) = max{x1 + 6, x2 + 7} s_A(x1) = max{12, x1 + 5} base order 2: precedence: plus# > s partial status: pi(plus#) = [1] 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: minus#(s(x),s(y)) -> minus#(x,y) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(|0|(),y) -> y r6: plus(s(x),y) -> s(plus(x,y)) r7: plus(s(x),y) -> plus(x,s(y)) r8: plus(s(x),y) -> s(plus(minus(x,y),double(y))) r9: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z)) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: minus#_A(x1,x2) = x1 + 2 s_A(x1) = max{11, x1 + 3} base order 2: precedence: minus# > s partial status: pi(minus#) = [1] 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: double#(s(x)) -> double#(x) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(|0|(),y) -> y r6: plus(s(x),y) -> s(plus(x,y)) r7: plus(s(x),y) -> plus(x,s(y)) r8: plus(s(x),y) -> s(plus(minus(x,y),double(y))) r9: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z)) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: double#_A(x1) = max{4, x1 + 1} s_A(x1) = max{12, x1 + 5} base order 2: precedence: s > double# partial status: pi(double#) = [1] pi(s) = [1] The following rules are removed: p1 Then no dependency pair remains.