YES
The TRS could be proven terminating. The proof took 1983 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (716ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (417ms).
| | – Problem 8 was processed with processor PolynomialLinearRange4iUR (176ms).
| | | – Problem 9 was processed with processor PolynomialLinearRange4iUR (120ms).
| | | | – Problem 10 was processed with processor PolynomialLinearRange4iUR (122ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor SubtermCriterion (36ms).
| | – Problem 7 was processed with processor SubtermCriterion (1ms).
| – Problem 5 was processed with processor SubtermCriterion (1ms).
| – Problem 6 was processed with processor SubtermCriterion (22ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| *#(*(x, y), z) | → | *#(x, *(y, z)) | | +#(0(x), 0(y)) | → | +#(x, y) |
| +#(+(x, y), z) | → | +#(y, z) | | *#(0(x), y) | → | 0#(*(x, y)) |
| sum#(cons(x, l)) | → | +#(x, sum(l)) | | prod#(cons(x, l)) | → | *#(x, prod(l)) |
| +#(+(x, y), z) | → | +#(x, +(y, z)) | | +#(0(x), 0(y)) | → | 0#(+(x, y)) |
| app#(cons(x, l1), l2) | → | app#(l1, l2) | | *#(x, +(y, z)) | → | +#(*(x, y), *(x, z)) |
| sum#(nil) | → | 0#(#) | | *#(x, +(y, z)) | → | *#(x, y) |
| prod#(app(l1, l2)) | → | prod#(l2) | | *#(0(x), y) | → | *#(x, y) |
| sum#(app(l1, l2)) | → | sum#(l2) | | +#(1(x), 0(y)) | → | +#(x, y) |
| +#(1(x), 1(y)) | → | 0#(+(+(x, y), 1(#))) | | +#(0(x), 1(y)) | → | +#(x, y) |
| sum#(app(l1, l2)) | → | +#(sum(l1), sum(l2)) | | *#(*(x, y), z) | → | *#(y, z) |
| sum#(cons(x, l)) | → | sum#(l) | | +#(1(x), 1(y)) | → | +#(+(x, y), 1(#)) |
| prod#(app(l1, l2)) | → | *#(prod(l1), prod(l2)) | | *#(1(x), y) | → | +#(0(*(x, y)), y) |
| *#(x, +(y, z)) | → | *#(x, z) | | prod#(cons(x, l)) | → | prod#(l) |
| *#(1(x), y) | → | *#(x, y) | | sum#(app(l1, l2)) | → | sum#(l1) |
| prod#(app(l1, l2)) | → | prod#(l1) | | *#(1(x), y) | → | 0#(*(x, y)) |
| +#(1(x), 1(y)) | → | +#(x, y) |
Rewrite Rules
| 0(#) | → | # | | +(x, #) | → | x |
| +(#, x) | → | x | | +(0(x), 0(y)) | → | 0(+(x, y)) |
| +(0(x), 1(y)) | → | 1(+(x, y)) | | +(1(x), 0(y)) | → | 1(+(x, y)) |
| +(1(x), 1(y)) | → | 0(+(+(x, y), 1(#))) | | +(+(x, y), z) | → | +(x, +(y, z)) |
| *(#, x) | → | # | | *(0(x), y) | → | 0(*(x, y)) |
| *(1(x), y) | → | +(0(*(x, y)), y) | | *(*(x, y), z) | → | *(x, *(y, z)) |
| *(x, +(y, z)) | → | +(*(x, y), *(x, z)) | | app(nil, l) | → | l |
| app(cons(x, l1), l2) | → | cons(x, app(l1, l2)) | | sum(nil) | → | 0(#) |
| sum(cons(x, l)) | → | +(x, sum(l)) | | sum(app(l1, l2)) | → | +(sum(l1), sum(l2)) |
| prod(nil) | → | 1(#) | | prod(cons(x, l)) | → | *(x, prod(l)) |
| prod(app(l1, l2)) | → | *(prod(l1), prod(l2)) |
Original Signature
Termination of terms over the following signature is verified: #, app, 1, 0, *, +, sum, nil, cons, prod
Strategy
The following SCCs where found
| sum#(app(l1, l2)) → sum#(l1) | sum#(cons(x, l)) → sum#(l) |
| sum#(app(l1, l2)) → sum#(l2) |
| app#(cons(x, l1), l2) → app#(l1, l2) |
| prod#(cons(x, l)) → prod#(l) | prod#(app(l1, l2)) → prod#(l1) |
| prod#(app(l1, l2)) → prod#(l2) |
| +#(0(x), 1(y)) → +#(x, y) | +#(0(x), 0(y)) → +#(x, y) |
| +#(+(x, y), z) → +#(y, z) | +#(1(x), 1(y)) → +#(x, y) |
| +#(1(x), 1(y)) → +#(+(x, y), 1(#)) | +#(1(x), 0(y)) → +#(x, y) |
| +#(+(x, y), z) → +#(x, +(y, z)) |
| *#(x, +(y, z)) → *#(x, z) | *#(1(x), y) → *#(x, y) |
| *#(*(x, y), z) → *#(y, z) | *#(*(x, y), z) → *#(x, *(y, z)) |
| *#(x, +(y, z)) → *#(x, y) | *#(0(x), y) → *#(x, y) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| +#(0(x), 1(y)) | → | +#(x, y) | | +#(0(x), 0(y)) | → | +#(x, y) |
| +#(+(x, y), z) | → | +#(y, z) | | +#(1(x), 1(y)) | → | +#(x, y) |
| +#(1(x), 1(y)) | → | +#(+(x, y), 1(#)) | | +#(1(x), 0(y)) | → | +#(x, y) |
| +#(+(x, y), z) | → | +#(x, +(y, z)) |
Rewrite Rules
| 0(#) | → | # | | +(x, #) | → | x |
| +(#, x) | → | x | | +(0(x), 0(y)) | → | 0(+(x, y)) |
| +(0(x), 1(y)) | → | 1(+(x, y)) | | +(1(x), 0(y)) | → | 1(+(x, y)) |
| +(1(x), 1(y)) | → | 0(+(+(x, y), 1(#))) | | +(+(x, y), z) | → | +(x, +(y, z)) |
| *(#, x) | → | # | | *(0(x), y) | → | 0(*(x, y)) |
| *(1(x), y) | → | +(0(*(x, y)), y) | | *(*(x, y), z) | → | *(x, *(y, z)) |
| *(x, +(y, z)) | → | +(*(x, y), *(x, z)) | | app(nil, l) | → | l |
| app(cons(x, l1), l2) | → | cons(x, app(l1, l2)) | | sum(nil) | → | 0(#) |
| sum(cons(x, l)) | → | +(x, sum(l)) | | sum(app(l1, l2)) | → | +(sum(l1), sum(l2)) |
| prod(nil) | → | 1(#) | | prod(cons(x, l)) | → | *(x, prod(l)) |
| prod(app(l1, l2)) | → | *(prod(l1), prod(l2)) |
Original Signature
Termination of terms over the following signature is verified: #, app, 1, 0, *, +, sum, nil, cons, prod
Strategy
Polynomial Interpretation
- #: 0
- *(x,y): 0
- +(x,y): y + x
- +#(x,y): y + 2x
- 0(x): 2x
- 1(x): 2x + 1
- app(x,y): 0
- cons(x,y): 0
- nil: 0
- prod(x): 0
- sum(x): 0
Improved Usable rules
| +(+(x, y), z) | → | +(x, +(y, z)) | | +(0(x), 0(y)) | → | 0(+(x, y)) |
| +(1(x), 0(y)) | → | 1(+(x, y)) | | 0(#) | → | # |
| +(x, #) | → | x | | +(0(x), 1(y)) | → | 1(+(x, y)) |
| +(1(x), 1(y)) | → | 0(+(+(x, y), 1(#))) | | +(#, x) | → | x |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| +#(0(x), 1(y)) | → | +#(x, y) | | +#(1(x), 1(y)) | → | +#(x, y) |
| +#(1(x), 1(y)) | → | +#(+(x, y), 1(#)) | | +#(1(x), 0(y)) | → | +#(x, y) |
Problem 8: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| +#(0(x), 0(y)) | → | +#(x, y) | | +#(+(x, y), z) | → | +#(y, z) |
| +#(+(x, y), z) | → | +#(x, +(y, z)) |
Rewrite Rules
| 0(#) | → | # | | +(x, #) | → | x |
| +(#, x) | → | x | | +(0(x), 0(y)) | → | 0(+(x, y)) |
| +(0(x), 1(y)) | → | 1(+(x, y)) | | +(1(x), 0(y)) | → | 1(+(x, y)) |
| +(1(x), 1(y)) | → | 0(+(+(x, y), 1(#))) | | +(+(x, y), z) | → | +(x, +(y, z)) |
| *(#, x) | → | # | | *(0(x), y) | → | 0(*(x, y)) |
| *(1(x), y) | → | +(0(*(x, y)), y) | | *(*(x, y), z) | → | *(x, *(y, z)) |
| *(x, +(y, z)) | → | +(*(x, y), *(x, z)) | | app(nil, l) | → | l |
| app(cons(x, l1), l2) | → | cons(x, app(l1, l2)) | | sum(nil) | → | 0(#) |
| sum(cons(x, l)) | → | +(x, sum(l)) | | sum(app(l1, l2)) | → | +(sum(l1), sum(l2)) |
| prod(nil) | → | 1(#) | | prod(cons(x, l)) | → | *(x, prod(l)) |
| prod(app(l1, l2)) | → | *(prod(l1), prod(l2)) |
Original Signature
Termination of terms over the following signature is verified: app, #, 1, 0, *, sum, +, prod, cons, nil
Strategy
Polynomial Interpretation
- #: 0
- *(x,y): 0
- +(x,y): y + x + 1
- +#(x,y): 2y + 2x
- 0(x): x
- 1(x): x + 1
- app(x,y): 0
- cons(x,y): 0
- nil: 0
- prod(x): 0
- sum(x): 0
Improved Usable rules
| +(+(x, y), z) | → | +(x, +(y, z)) | | +(0(x), 0(y)) | → | 0(+(x, y)) |
| +(1(x), 0(y)) | → | 1(+(x, y)) | | 0(#) | → | # |
| +(x, #) | → | x | | +(0(x), 1(y)) | → | 1(+(x, y)) |
| +(1(x), 1(y)) | → | 0(+(+(x, y), 1(#))) | | +(#, x) | → | x |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| +#(+(x, y), z) | → | +#(y, z) |
Problem 9: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| +#(0(x), 0(y)) | → | +#(x, y) | | +#(+(x, y), z) | → | +#(x, +(y, z)) |
Rewrite Rules
| 0(#) | → | # | | +(x, #) | → | x |
| +(#, x) | → | x | | +(0(x), 0(y)) | → | 0(+(x, y)) |
| +(0(x), 1(y)) | → | 1(+(x, y)) | | +(1(x), 0(y)) | → | 1(+(x, y)) |
| +(1(x), 1(y)) | → | 0(+(+(x, y), 1(#))) | | +(+(x, y), z) | → | +(x, +(y, z)) |
| *(#, x) | → | # | | *(0(x), y) | → | 0(*(x, y)) |
| *(1(x), y) | → | +(0(*(x, y)), y) | | *(*(x, y), z) | → | *(x, *(y, z)) |
| *(x, +(y, z)) | → | +(*(x, y), *(x, z)) | | app(nil, l) | → | l |
| app(cons(x, l1), l2) | → | cons(x, app(l1, l2)) | | sum(nil) | → | 0(#) |
| sum(cons(x, l)) | → | +(x, sum(l)) | | sum(app(l1, l2)) | → | +(sum(l1), sum(l2)) |
| prod(nil) | → | 1(#) | | prod(cons(x, l)) | → | *(x, prod(l)) |
| prod(app(l1, l2)) | → | *(prod(l1), prod(l2)) |
Original Signature
Termination of terms over the following signature is verified: #, app, 1, 0, *, +, sum, nil, cons, prod
Strategy
Polynomial Interpretation
- #: 0
- *(x,y): 0
- +(x,y): 2x
- +#(x,y): 2x
- 0(x): x + 1
- 1(x): 1
- app(x,y): 0
- cons(x,y): 0
- nil: 0
- prod(x): 0
- sum(x): 0
Improved Usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| +#(0(x), 0(y)) | → | +#(x, y) |
Problem 10: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| +#(+(x, y), z) | → | +#(x, +(y, z)) |
Rewrite Rules
| 0(#) | → | # | | +(x, #) | → | x |
| +(#, x) | → | x | | +(0(x), 0(y)) | → | 0(+(x, y)) |
| +(0(x), 1(y)) | → | 1(+(x, y)) | | +(1(x), 0(y)) | → | 1(+(x, y)) |
| +(1(x), 1(y)) | → | 0(+(+(x, y), 1(#))) | | +(+(x, y), z) | → | +(x, +(y, z)) |
| *(#, x) | → | # | | *(0(x), y) | → | 0(*(x, y)) |
| *(1(x), y) | → | +(0(*(x, y)), y) | | *(*(x, y), z) | → | *(x, *(y, z)) |
| *(x, +(y, z)) | → | +(*(x, y), *(x, z)) | | app(nil, l) | → | l |
| app(cons(x, l1), l2) | → | cons(x, app(l1, l2)) | | sum(nil) | → | 0(#) |
| sum(cons(x, l)) | → | +(x, sum(l)) | | sum(app(l1, l2)) | → | +(sum(l1), sum(l2)) |
| prod(nil) | → | 1(#) | | prod(cons(x, l)) | → | *(x, prod(l)) |
| prod(app(l1, l2)) | → | *(prod(l1), prod(l2)) |
Original Signature
Termination of terms over the following signature is verified: app, #, 1, 0, *, sum, +, prod, cons, nil
Strategy
Polynomial Interpretation
- #: 1
- *(x,y): 0
- +(x,y): y + x + 1
- +#(x,y): x + 1
- 0(x): 0
- 1(x): 0
- app(x,y): 0
- cons(x,y): 0
- nil: 0
- prod(x): 0
- sum(x): 0
Improved Usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| +#(+(x, y), z) | → | +#(x, +(y, z)) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| prod#(cons(x, l)) | → | prod#(l) | | prod#(app(l1, l2)) | → | prod#(l1) |
| prod#(app(l1, l2)) | → | prod#(l2) |
Rewrite Rules
| 0(#) | → | # | | +(x, #) | → | x |
| +(#, x) | → | x | | +(0(x), 0(y)) | → | 0(+(x, y)) |
| +(0(x), 1(y)) | → | 1(+(x, y)) | | +(1(x), 0(y)) | → | 1(+(x, y)) |
| +(1(x), 1(y)) | → | 0(+(+(x, y), 1(#))) | | +(+(x, y), z) | → | +(x, +(y, z)) |
| *(#, x) | → | # | | *(0(x), y) | → | 0(*(x, y)) |
| *(1(x), y) | → | +(0(*(x, y)), y) | | *(*(x, y), z) | → | *(x, *(y, z)) |
| *(x, +(y, z)) | → | +(*(x, y), *(x, z)) | | app(nil, l) | → | l |
| app(cons(x, l1), l2) | → | cons(x, app(l1, l2)) | | sum(nil) | → | 0(#) |
| sum(cons(x, l)) | → | +(x, sum(l)) | | sum(app(l1, l2)) | → | +(sum(l1), sum(l2)) |
| prod(nil) | → | 1(#) | | prod(cons(x, l)) | → | *(x, prod(l)) |
| prod(app(l1, l2)) | → | *(prod(l1), prod(l2)) |
Original Signature
Termination of terms over the following signature is verified: #, app, 1, 0, *, +, sum, nil, cons, prod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| prod#(cons(x, l)) | → | prod#(l) | | prod#(app(l1, l2)) | → | prod#(l1) |
| prod#(app(l1, l2)) | → | prod#(l2) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| *#(x, +(y, z)) | → | *#(x, z) | | *#(1(x), y) | → | *#(x, y) |
| *#(*(x, y), z) | → | *#(y, z) | | *#(*(x, y), z) | → | *#(x, *(y, z)) |
| *#(x, +(y, z)) | → | *#(x, y) | | *#(0(x), y) | → | *#(x, y) |
Rewrite Rules
| 0(#) | → | # | | +(x, #) | → | x |
| +(#, x) | → | x | | +(0(x), 0(y)) | → | 0(+(x, y)) |
| +(0(x), 1(y)) | → | 1(+(x, y)) | | +(1(x), 0(y)) | → | 1(+(x, y)) |
| +(1(x), 1(y)) | → | 0(+(+(x, y), 1(#))) | | +(+(x, y), z) | → | +(x, +(y, z)) |
| *(#, x) | → | # | | *(0(x), y) | → | 0(*(x, y)) |
| *(1(x), y) | → | +(0(*(x, y)), y) | | *(*(x, y), z) | → | *(x, *(y, z)) |
| *(x, +(y, z)) | → | +(*(x, y), *(x, z)) | | app(nil, l) | → | l |
| app(cons(x, l1), l2) | → | cons(x, app(l1, l2)) | | sum(nil) | → | 0(#) |
| sum(cons(x, l)) | → | +(x, sum(l)) | | sum(app(l1, l2)) | → | +(sum(l1), sum(l2)) |
| prod(nil) | → | 1(#) | | prod(cons(x, l)) | → | *(x, prod(l)) |
| prod(app(l1, l2)) | → | *(prod(l1), prod(l2)) |
Original Signature
Termination of terms over the following signature is verified: #, app, 1, 0, *, +, sum, nil, cons, prod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| *#(1(x), y) | → | *#(x, y) | | *#(*(x, y), z) | → | *#(x, *(y, z)) |
| *#(*(x, y), z) | → | *#(y, z) | | *#(0(x), y) | → | *#(x, y) |
Problem 7: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| *#(x, +(y, z)) | → | *#(x, z) | | *#(x, +(y, z)) | → | *#(x, y) |
Rewrite Rules
| 0(#) | → | # | | +(x, #) | → | x |
| +(#, x) | → | x | | +(0(x), 0(y)) | → | 0(+(x, y)) |
| +(0(x), 1(y)) | → | 1(+(x, y)) | | +(1(x), 0(y)) | → | 1(+(x, y)) |
| +(1(x), 1(y)) | → | 0(+(+(x, y), 1(#))) | | +(+(x, y), z) | → | +(x, +(y, z)) |
| *(#, x) | → | # | | *(0(x), y) | → | 0(*(x, y)) |
| *(1(x), y) | → | +(0(*(x, y)), y) | | *(*(x, y), z) | → | *(x, *(y, z)) |
| *(x, +(y, z)) | → | +(*(x, y), *(x, z)) | | app(nil, l) | → | l |
| app(cons(x, l1), l2) | → | cons(x, app(l1, l2)) | | sum(nil) | → | 0(#) |
| sum(cons(x, l)) | → | +(x, sum(l)) | | sum(app(l1, l2)) | → | +(sum(l1), sum(l2)) |
| prod(nil) | → | 1(#) | | prod(cons(x, l)) | → | *(x, prod(l)) |
| prod(app(l1, l2)) | → | *(prod(l1), prod(l2)) |
Original Signature
Termination of terms over the following signature is verified: app, #, 1, 0, *, sum, +, prod, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| *#(x, +(y, z)) | → | *#(x, z) | | *#(x, +(y, z)) | → | *#(x, y) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| app#(cons(x, l1), l2) | → | app#(l1, l2) |
Rewrite Rules
| 0(#) | → | # | | +(x, #) | → | x |
| +(#, x) | → | x | | +(0(x), 0(y)) | → | 0(+(x, y)) |
| +(0(x), 1(y)) | → | 1(+(x, y)) | | +(1(x), 0(y)) | → | 1(+(x, y)) |
| +(1(x), 1(y)) | → | 0(+(+(x, y), 1(#))) | | +(+(x, y), z) | → | +(x, +(y, z)) |
| *(#, x) | → | # | | *(0(x), y) | → | 0(*(x, y)) |
| *(1(x), y) | → | +(0(*(x, y)), y) | | *(*(x, y), z) | → | *(x, *(y, z)) |
| *(x, +(y, z)) | → | +(*(x, y), *(x, z)) | | app(nil, l) | → | l |
| app(cons(x, l1), l2) | → | cons(x, app(l1, l2)) | | sum(nil) | → | 0(#) |
| sum(cons(x, l)) | → | +(x, sum(l)) | | sum(app(l1, l2)) | → | +(sum(l1), sum(l2)) |
| prod(nil) | → | 1(#) | | prod(cons(x, l)) | → | *(x, prod(l)) |
| prod(app(l1, l2)) | → | *(prod(l1), prod(l2)) |
Original Signature
Termination of terms over the following signature is verified: #, app, 1, 0, *, +, sum, nil, cons, prod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| app#(cons(x, l1), l2) | → | app#(l1, l2) |
Problem 6: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sum#(app(l1, l2)) | → | sum#(l1) | | sum#(cons(x, l)) | → | sum#(l) |
| sum#(app(l1, l2)) | → | sum#(l2) |
Rewrite Rules
| 0(#) | → | # | | +(x, #) | → | x |
| +(#, x) | → | x | | +(0(x), 0(y)) | → | 0(+(x, y)) |
| +(0(x), 1(y)) | → | 1(+(x, y)) | | +(1(x), 0(y)) | → | 1(+(x, y)) |
| +(1(x), 1(y)) | → | 0(+(+(x, y), 1(#))) | | +(+(x, y), z) | → | +(x, +(y, z)) |
| *(#, x) | → | # | | *(0(x), y) | → | 0(*(x, y)) |
| *(1(x), y) | → | +(0(*(x, y)), y) | | *(*(x, y), z) | → | *(x, *(y, z)) |
| *(x, +(y, z)) | → | +(*(x, y), *(x, z)) | | app(nil, l) | → | l |
| app(cons(x, l1), l2) | → | cons(x, app(l1, l2)) | | sum(nil) | → | 0(#) |
| sum(cons(x, l)) | → | +(x, sum(l)) | | sum(app(l1, l2)) | → | +(sum(l1), sum(l2)) |
| prod(nil) | → | 1(#) | | prod(cons(x, l)) | → | *(x, prod(l)) |
| prod(app(l1, l2)) | → | *(prod(l1), prod(l2)) |
Original Signature
Termination of terms over the following signature is verified: #, app, 1, 0, *, +, sum, nil, cons, prod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| sum#(app(l1, l2)) | → | sum#(l1) | | sum#(cons(x, l)) | → | sum#(l) |
| sum#(app(l1, l2)) | → | sum#(l2) |