The rewrite relation of the following TRS is considered.
p(a(x0),p(a(b(x1)),x2)) | → | p(a(b(a(x2))),p(a(a(x1)),x2)) | (1) |
p#(a(x0),p(a(b(x1)),x2)) | → | p#(a(a(x1)),x2) | (2) |
p#(a(x0),p(a(b(x1)),x2)) | → | p#(a(b(a(x2))),p(a(a(x1)),x2)) | (3) |
The dependency pairs are split into 1 component.
p#(a(x0),p(a(b(x1)),x2)) | → | p#(a(b(a(x2))),p(a(a(x1)),x2)) | (3) |
p#(a(x0),p(a(b(x1)),x2)) | → | p#(a(a(x1)),x2) | (2) |
[a(x1)] | = | 1 |
[b(x1)] | = | x1 + 1 |
[p#(x1, x2)] | = | x2 + 0 |
[p(x1, x2)] | = | x1 + x2 + 38 |
p(a(x0),p(a(b(x1)),x2)) | → | p(a(b(a(x2))),p(a(a(x1)),x2)) | (1) |
p#(a(x0),p(a(b(x1)),x2)) | → | p#(a(a(x1)),x2) | (2) |
The dependency pairs are split into 1 component.
p#(a(x0),p(a(b(x1)),x2)) | → | p#(a(b(a(x2))),p(a(a(x1)),x2)) | (3) |
π(a) | = | 1 |
prec(b) | = | 2 | status(b) | = | [1] | list-extension(b) | = | Lex | ||
prec(p#) | = | 1 | status(p#) | = | [1, 2] | list-extension(p#) | = | Lex | ||
prec(p) | = | 2 | status(p) | = | [2, 1] | list-extension(p) | = | Lex |
[b(x1)] | = | x1 + 0 |
[p#(x1, x2)] | = | max(x1 + 0, x2 + 0, 0) |
[p(x1, x2)] | = | max(x1 + 0, x2 + 0, 0) |
p(a(x0),p(a(b(x1)),x2)) | → | p(a(b(a(x2))),p(a(a(x1)),x2)) | (1) |
p#(a(x0),p(a(b(x1)),x2)) | → | p#(a(b(a(x2))),p(a(a(x1)),x2)) | (3) |
The dependency pairs are split into 0 components.