YES
The TRS could be proven terminating. The proof took 83 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (26ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
| – Problem 5 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| prod#(cons(x, l)) | → | prod#(l) | | *#(s(x), s(y)) | → | *#(x, y) |
| *#(s(x), s(y)) | → | +#(x, y) | | *#(s(x), s(y)) | → | +#(*(x, y), +(x, y)) |
| sum#(cons(x, l)) | → | sum#(l) | | sum#(cons(x, l)) | → | +#(x, sum(l)) |
| prod#(cons(x, l)) | → | *#(x, prod(l)) | | +#(s(x), s(y)) | → | +#(x, y) |
Rewrite Rules
| +(x, 0) | → | x | | +(0, x) | → | x |
| +(s(x), s(y)) | → | s(s(+(x, y))) | | *(x, 0) | → | 0 |
| *(0, x) | → | 0 | | *(s(x), s(y)) | → | s(+(*(x, y), +(x, y))) |
| sum(nil) | → | 0 | | sum(cons(x, l)) | → | +(x, sum(l)) |
| prod(nil) | → | s(0) | | prod(cons(x, l)) | → | *(x, prod(l)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, *, +, sum, nil, cons, prod
Strategy
The following SCCs where found
| prod#(cons(x, l)) → prod#(l) |
| *#(s(x), s(y)) → *#(x, y) |
| sum#(cons(x, l)) → sum#(l) |
| +#(s(x), s(y)) → +#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sum#(cons(x, l)) | → | sum#(l) |
Rewrite Rules
| +(x, 0) | → | x | | +(0, x) | → | x |
| +(s(x), s(y)) | → | s(s(+(x, y))) | | *(x, 0) | → | 0 |
| *(0, x) | → | 0 | | *(s(x), s(y)) | → | s(+(*(x, y), +(x, y))) |
| sum(nil) | → | 0 | | sum(cons(x, l)) | → | +(x, sum(l)) |
| prod(nil) | → | s(0) | | prod(cons(x, l)) | → | *(x, prod(l)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, *, +, sum, nil, cons, prod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| sum#(cons(x, l)) | → | sum#(l) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| prod#(cons(x, l)) | → | prod#(l) |
Rewrite Rules
| +(x, 0) | → | x | | +(0, x) | → | x |
| +(s(x), s(y)) | → | s(s(+(x, y))) | | *(x, 0) | → | 0 |
| *(0, x) | → | 0 | | *(s(x), s(y)) | → | s(+(*(x, y), +(x, y))) |
| sum(nil) | → | 0 | | sum(cons(x, l)) | → | +(x, sum(l)) |
| prod(nil) | → | s(0) | | prod(cons(x, l)) | → | *(x, prod(l)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, *, +, sum, nil, cons, prod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| prod#(cons(x, l)) | → | prod#(l) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| *#(s(x), s(y)) | → | *#(x, y) |
Rewrite Rules
| +(x, 0) | → | x | | +(0, x) | → | x |
| +(s(x), s(y)) | → | s(s(+(x, y))) | | *(x, 0) | → | 0 |
| *(0, x) | → | 0 | | *(s(x), s(y)) | → | s(+(*(x, y), +(x, y))) |
| sum(nil) | → | 0 | | sum(cons(x, l)) | → | +(x, sum(l)) |
| prod(nil) | → | s(0) | | prod(cons(x, l)) | → | *(x, prod(l)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, *, +, sum, nil, cons, prod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| *#(s(x), s(y)) | → | *#(x, y) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| +#(s(x), s(y)) | → | +#(x, y) |
Rewrite Rules
| +(x, 0) | → | x | | +(0, x) | → | x |
| +(s(x), s(y)) | → | s(s(+(x, y))) | | *(x, 0) | → | 0 |
| *(0, x) | → | 0 | | *(s(x), s(y)) | → | s(+(*(x, y), +(x, y))) |
| sum(nil) | → | 0 | | sum(cons(x, l)) | → | +(x, sum(l)) |
| prod(nil) | → | s(0) | | prod(cons(x, l)) | → | *(x, prod(l)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, *, +, sum, nil, cons, prod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| +#(s(x), s(y)) | → | +#(x, y) |