The rewrite relation of the following TRS is considered.
p(0(x1)) | → | s(s(0(s(s(p(x1)))))) | (1) |
p(s(0(x1))) | → | 0(x1) | (2) |
p(s(s(x1))) | → | s(p(s(x1))) | (3) |
f(s(x1)) | → | g(q(i(x1))) | (4) |
g(x1) | → | f(p(p(x1))) | (5) |
q(i(x1)) | → | q(s(x1)) | (6) |
q(s(x1)) | → | s(s(x1)) | (7) |
i(x1) | → | s(x1) | (8) |
0(p(x1)) | → | p(s(s(0(s(s(x1)))))) | (9) |
0(s(p(x1))) | → | 0(x1) | (10) |
s(s(p(x1))) | → | s(p(s(x1))) | (11) |
s(f(x1)) | → | i(q(g(x1))) | (12) |
g(x1) | → | p(p(f(x1))) | (13) |
i(q(x1)) | → | s(q(x1)) | (14) |
s(q(x1)) | → | s(s(x1)) | (15) |
i(x1) | → | s(x1) | (8) |
The TRS is overlay and locally confluent:
10Hence, it suffices to show innermost termination in the following.
0#(p(x1)) | → | s#(s(0(s(s(x1))))) | (16) |
0#(p(x1)) | → | s#(0(s(s(x1)))) | (17) |
0#(p(x1)) | → | 0#(s(s(x1))) | (18) |
0#(p(x1)) | → | s#(s(x1)) | (19) |
0#(p(x1)) | → | s#(x1) | (20) |
0#(s(p(x1))) | → | 0#(x1) | (21) |
s#(s(p(x1))) | → | s#(p(s(x1))) | (22) |
s#(s(p(x1))) | → | s#(x1) | (23) |
s#(f(x1)) | → | i#(q(g(x1))) | (24) |
s#(f(x1)) | → | g#(x1) | (25) |
i#(q(x1)) | → | s#(q(x1)) | (26) |
s#(q(x1)) | → | s#(s(x1)) | (27) |
s#(q(x1)) | → | s#(x1) | (28) |
i#(x1) | → | s#(x1) | (29) |
The dependency pairs are split into 2 components.
0#(s(p(x1))) | → | 0#(x1) | (21) |
0#(p(x1)) | → | 0#(s(s(x1))) | (18) |
We restrict the rewrite rules to the following usable rules of the DP problem.
s(s(p(x1))) | → | s(p(s(x1))) | (11) |
s(f(x1)) | → | i(q(g(x1))) | (12) |
i(q(x1)) | → | s(q(x1)) | (14) |
s(q(x1)) | → | s(s(x1)) | (15) |
i(x1) | → | s(x1) | (8) |
g(x1) | → | p(p(f(x1))) | (13) |
We restrict the innermost strategy to the following left hand sides.
s(s(p(x0))) |
s(f(x0)) |
g(x0) |
s(q(x0)) |
i(x0) |
[0#(x1)] | = |
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[s(x1)] | = |
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[p(x1)] | = |
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[f(x1)] | = |
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[i(x1)] | = |
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[q(x1)] | = |
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[g(x1)] | = |
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0#(p(x1)) | → | 0#(s(s(x1))) | (18) |
We restrict the rewrite rules to the following usable rules of the DP problem.
There are no rules.
We restrict the innermost strategy to the following left hand sides.
s(s(p(x0))) |
s(f(x0)) |
s(q(x0)) |
Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.
0#(s(p(x1))) | → | 0#(x1) | (21) |
1 | > | 1 |
As there is no critical graph in the transitive closure, there are no infinite chains.
s#(f(x1)) | → | i#(q(g(x1))) | (24) |
i#(q(x1)) | → | s#(q(x1)) | (26) |
s#(q(x1)) | → | s#(s(x1)) | (27) |
s#(s(p(x1))) | → | s#(x1) | (23) |
s#(q(x1)) | → | s#(x1) | (28) |
i#(x1) | → | s#(x1) | (29) |
We restrict the rewrite rules to the following usable rules of the DP problem.
s(s(p(x1))) | → | s(p(s(x1))) | (11) |
s(f(x1)) | → | i(q(g(x1))) | (12) |
i(q(x1)) | → | s(q(x1)) | (14) |
s(q(x1)) | → | s(s(x1)) | (15) |
i(x1) | → | s(x1) | (8) |
g(x1) | → | p(p(f(x1))) | (13) |
We restrict the innermost strategy to the following left hand sides.
s(s(p(x0))) |
s(f(x0)) |
g(x0) |
s(q(x0)) |
i(x0) |
[i#(x1)] | = | 2 + 2 · x1 |
[q(x1)] | = | 2 + x1 |
[s#(x1)] | = | 2 + 2 · x1 |
[i(x1)] | = | 2 + x1 |
[g(x1)] | = | 0 |
[p(x1)] | = | -2 + x1 |
[f(x1)] | = | 2 |
[s(x1)] | = | 2 + x1 |
s#(q(x1)) | → | s#(x1) | (28) |
[i#(x1)] | = | 2 · x1 |
[q(x1)] | = | 2 + 2 · x1 |
[s#(x1)] | = | 2 · x1 |
[i(x1)] | = | 2 + x1 |
[g(x1)] | = | 0 |
[p(x1)] | = | -1 + x1 |
[f(x1)] | = | 2 |
[s(x1)] | = | 2 + x1 |
s#(s(p(x1))) | → | s#(x1) | (23) |
The dependency pairs are split into 1 component.
i#(x1) | → | s#(x1) | (29) |
s#(f(x1)) | → | i#(q(g(x1))) | (24) |
We restrict the rewrite rules to the following usable rules of the DP problem.
g(x1) | → | p(p(f(x1))) | (13) |
We restrict the innermost strategy to the following left hand sides.
g(x0) |
[i#(x1)] | = | 1 + 2 · x1 |
[q(x1)] | = | 1 |
[g(x1)] | = | 2 + 2 · x1 |
[p(x1)] | = | -2 |
[f(x1)] | = | 2 |
[s#(x1)] | = | 2 · x1 |
i#(x1) | → | s#(x1) | (29) |
s#(f(x1)) | → | i#(q(g(x1))) | (24) |
There are no pairs anymore.