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))) -- 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) 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))) The estimated dependency graph contains the following SCCs: {p3, p4, p5} {p1} {p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: plus#(s(x),y) -> plus#(minus(x,y),double(y)) p2: plus#(s(x),y) -> plus#(x,s(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))) The set of usable rules consists of r1, r2 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 - 4} s_A(x1) = x1 + 16 minus_A(x1,x2) = max{3, x1} double_A(x1) = max{13, x1 - 7} |0|_A = 0 base order 2: precedence: s = double > plus# > minus = |0| partial status: pi(plus#) = [] pi(s) = [1] pi(minus) = [1] pi(double) = [] pi(|0|) = [] 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,s(y)) 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))) 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(x),y) -> plus#(x,s(y)) 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))) 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{1, x1 - 4} s_A(x1) = x1 + 6 base order 2: precedence: s > plus# partial status: pi(plus#) = [] pi(s) = [] The following rules are removed: p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: plus#(s(x),y) -> plus#(x,s(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))) 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,s(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))) 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) = x1 + 1 s_A(x1) = max{4, x1 + 2} base order 2: precedence: plus# > s partial status: pi(plus#) = [1] pi(s) = [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: 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))) 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) = max{x1 - 1, x2 + 5} s_A(x1) = max{7, x1 + 4} base order 2: precedence: minus# > s partial status: pi(minus#) = [2] pi(s) = [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: 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))) 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{8, x1 + 2} s_A(x1) = max{10, x1 + 7} 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.