The rewrite relation of the following TRS is considered.
c(c(z,y,a),a,a) | → | b(z,y) | (1) |
f(c(x,y,z)) | → | c(z,f(b(y,z)),a) | (2) |
b(z,b(c(a,y,a),f(f(x)))) | → | c(c(y,a,z),z,x) | (3) |
f#(c(x,y,z)) | → | f#(b(y,z)) | (4) |
c#(c(z,y,a),a,a) | → | b#(z,y) | (5) |
b#(z,b(c(a,y,a),f(f(x)))) | → | c#(y,a,z) | (6) |
f#(c(x,y,z)) | → | b#(y,z) | (7) |
f#(c(x,y,z)) | → | c#(z,f(b(y,z)),a) | (8) |
b#(z,b(c(a,y,a),f(f(x)))) | → | c#(c(y,a,z),z,x) | (9) |
The dependency pairs are split into 2 components.
f#(c(x,y,z)) | → | f#(b(y,z)) | (4) |
prec(a) | = | 1 | status(a) | = | [] | list-extension(a) | = | Lex | ||
prec(b) | = | 1 | status(b) | = | [1, 2] | list-extension(b) | = | Lex | ||
prec(c) | = | 1 | status(c) | = | [2, 3, 1] | list-extension(c) | = | Lex | ||
prec(f) | = | 0 | status(f) | = | [] | list-extension(f) | = | Lex | ||
prec(f#) | = | 0 | status(f#) | = | [1] | list-extension(f#) | = | Lex | ||
prec(c#) | = | 0 | status(c#) | = | [] | list-extension(c#) | = | Lex | ||
prec(b#) | = | 0 | status(b#) | = | [1, 2] | list-extension(b#) | = | Lex |
[a] | = | 32287 |
[b(x1, x2)] | = | max(x1 + 161442, x2 + 32288, 0) |
[c(x1, x2, x3)] | = | max(x1 + 129153, x2 + 161442, x3 + 32288, 0) |
[f(x1)] | = | x1 + 64576 |
[f#(x1)] | = | x1 + 1 |
[c#(x1, x2, x3)] | = | max(x1 + 1, 0) |
[b#(x1, x2)] | = | max(x1 + 1, x2 + 1, 0) |
c(c(z,y,a),a,a) | → | b(z,y) | (1) |
b(z,b(c(a,y,a),f(f(x)))) | → | c(c(y,a,z),z,x) | (3) |
f#(c(x,y,z)) | → | f#(b(y,z)) | (4) |
The dependency pairs are split into 0 components.
b#(z,b(c(a,y,a),f(f(x)))) | → | c#(c(y,a,z),z,x) | (9) |
c#(c(z,y,a),a,a) | → | b#(z,y) | (5) |
b#(z,b(c(a,y,a),f(f(x)))) | → | c#(y,a,z) | (6) |
[a] | = | 1 |
[b(x1, x2)] | = | x1 + x2 + 0 |
[c(x1, x2, x3)] | = | x1 + x2 + 1 |
[f(x1)] | = | x1 + 1 |
[f#(x1)] | = | 0 |
[c#(x1, x2, x3)] | = | x1 + x2 + x3 + 0 |
[b#(x1, x2)] | = | x1 + x2 + 0 |
c(c(z,y,a),a,a) | → | b(z,y) | (1) |
b(z,b(c(a,y,a),f(f(x)))) | → | c(c(y,a,z),z,x) | (3) |
b#(z,b(c(a,y,a),f(f(x)))) | → | c#(c(y,a,z),z,x) | (9) |
c#(c(z,y,a),a,a) | → | b#(z,y) | (5) |
b#(z,b(c(a,y,a),f(f(x)))) | → | c#(y,a,z) | (6) |
The dependency pairs are split into 0 components.