TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60001 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (117ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| | – Problem 6 was processed with processor PolynomialLinearRange4iUR (35ms).
| – Problem 4 was processed with processor PolynomialLinearRange4iUR (100ms).
| | – Problem 7 was processed with processor PolynomialLinearRange4iUR (21ms).
| – Problem 5 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (1072ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (1016ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (1008ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (20498ms), DependencyGraph (0ms), ReductionPairSAT (5353ms), DependencyGraph (1ms), SizeChangePrinciple (738ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (3ms)].
The following open problems remain:
Open Dependency Pair Problem 5
Dependency Pairs
| quot#(s(x), s(y), z) | → | quot#(minus(p(ack(0, x)), y), s(y), s(z)) |
Rewrite Rules
| minus(minus(x, y), z) | → | minus(x, plus(y, z)) | | minus(0, y) | → | 0 |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| plus(0, y) | → | y | | plus(s(x), y) | → | plus(x, s(y)) |
| plus(s(x), y) | → | s(plus(y, x)) | | zero(s(x)) | → | false |
| zero(0) | → | true | | p(s(x)) | → | x |
| p(0) | → | 0 | | div(x, y) | → | quot(x, y, 0) |
| quot(s(x), s(y), z) | → | quot(minus(p(ack(0, x)), y), s(y), s(z)) | | quot(0, s(y), z) | → | z |
| ack(0, x) | → | s(x) | | ack(0, x) | → | plus(x, s(0)) |
| ack(s(x), 0) | → | ack(x, s(0)) | | ack(s(x), s(y)) | → | ack(x, ack(s(x), y)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, minus, s, ack, p, div, true, false, zero, quot
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| div#(x, y) | → | quot#(x, y, 0) | | minus#(minus(x, y), z) | → | plus#(y, z) |
| ack#(0, x) | → | plus#(x, s(0)) | | ack#(s(x), s(y)) | → | ack#(s(x), y) |
| quot#(s(x), s(y), z) | → | ack#(0, x) | | quot#(s(x), s(y), z) | → | minus#(p(ack(0, x)), y) |
| minus#(minus(x, y), z) | → | minus#(x, plus(y, z)) | | quot#(s(x), s(y), z) | → | quot#(minus(p(ack(0, x)), y), s(y), s(z)) |
| plus#(s(x), y) | → | plus#(y, x) | | ack#(s(x), s(y)) | → | ack#(x, ack(s(x), y)) |
| minus#(s(x), s(y)) | → | minus#(x, y) | | ack#(s(x), 0) | → | ack#(x, s(0)) |
| quot#(s(x), s(y), z) | → | p#(ack(0, x)) | | plus#(s(x), y) | → | plus#(x, s(y)) |
Rewrite Rules
| minus(minus(x, y), z) | → | minus(x, plus(y, z)) | | minus(0, y) | → | 0 |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| plus(0, y) | → | y | | plus(s(x), y) | → | plus(x, s(y)) |
| plus(s(x), y) | → | s(plus(y, x)) | | zero(s(x)) | → | false |
| zero(0) | → | true | | p(s(x)) | → | x |
| p(0) | → | 0 | | div(x, y) | → | quot(x, y, 0) |
| quot(s(x), s(y), z) | → | quot(minus(p(ack(0, x)), y), s(y), s(z)) | | quot(0, s(y), z) | → | z |
| ack(0, x) | → | s(x) | | ack(0, x) | → | plus(x, s(0)) |
| ack(s(x), 0) | → | ack(x, s(0)) | | ack(s(x), s(y)) | → | ack(x, ack(s(x), y)) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, ack, p, div, false, true, zero, quot
Strategy
The following SCCs where found
| minus#(s(x), s(y)) → minus#(x, y) | minus#(minus(x, y), z) → minus#(x, plus(y, z)) |
| quot#(s(x), s(y), z) → quot#(minus(p(ack(0, x)), y), s(y), s(z)) |
| ack#(s(x), s(y)) → ack#(x, ack(s(x), y)) | ack#(s(x), s(y)) → ack#(s(x), y) |
| ack#(s(x), 0) → ack#(x, s(0)) |
| plus#(s(x), y) → plus#(y, x) | plus#(s(x), y) → plus#(x, s(y)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(s(x), s(y)) | → | minus#(x, y) | | minus#(minus(x, y), z) | → | minus#(x, plus(y, z)) |
Rewrite Rules
| minus(minus(x, y), z) | → | minus(x, plus(y, z)) | | minus(0, y) | → | 0 |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| plus(0, y) | → | y | | plus(s(x), y) | → | plus(x, s(y)) |
| plus(s(x), y) | → | s(plus(y, x)) | | zero(s(x)) | → | false |
| zero(0) | → | true | | p(s(x)) | → | x |
| p(0) | → | 0 | | div(x, y) | → | quot(x, y, 0) |
| quot(s(x), s(y), z) | → | quot(minus(p(ack(0, x)), y), s(y), s(z)) | | quot(0, s(y), z) | → | z |
| ack(0, x) | → | s(x) | | ack(0, x) | → | plus(x, s(0)) |
| ack(s(x), 0) | → | ack(x, s(0)) | | ack(s(x), s(y)) | → | ack(x, ack(s(x), y)) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, ack, p, div, false, true, zero, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(x), s(y)) | → | minus#(x, y) | | minus#(minus(x, y), z) | → | minus#(x, plus(y, z)) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| ack#(s(x), s(y)) | → | ack#(x, ack(s(x), y)) | | ack#(s(x), s(y)) | → | ack#(s(x), y) |
| ack#(s(x), 0) | → | ack#(x, s(0)) |
Rewrite Rules
| minus(minus(x, y), z) | → | minus(x, plus(y, z)) | | minus(0, y) | → | 0 |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| plus(0, y) | → | y | | plus(s(x), y) | → | plus(x, s(y)) |
| plus(s(x), y) | → | s(plus(y, x)) | | zero(s(x)) | → | false |
| zero(0) | → | true | | p(s(x)) | → | x |
| p(0) | → | 0 | | div(x, y) | → | quot(x, y, 0) |
| quot(s(x), s(y), z) | → | quot(minus(p(ack(0, x)), y), s(y), s(z)) | | quot(0, s(y), z) | → | z |
| ack(0, x) | → | s(x) | | ack(0, x) | → | plus(x, s(0)) |
| ack(s(x), 0) | → | ack(x, s(0)) | | ack(s(x), s(y)) | → | ack(x, ack(s(x), y)) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, ack, p, div, false, true, zero, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| ack#(s(x), s(y)) | → | ack#(x, ack(s(x), y)) | | ack#(s(x), 0) | → | ack#(x, s(0)) |
Problem 6: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| ack#(s(x), s(y)) | → | ack#(s(x), y) |
Rewrite Rules
| minus(minus(x, y), z) | → | minus(x, plus(y, z)) | | minus(0, y) | → | 0 |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| plus(0, y) | → | y | | plus(s(x), y) | → | plus(x, s(y)) |
| plus(s(x), y) | → | s(plus(y, x)) | | zero(s(x)) | → | false |
| zero(0) | → | true | | p(s(x)) | → | x |
| p(0) | → | 0 | | div(x, y) | → | quot(x, y, 0) |
| quot(s(x), s(y), z) | → | quot(minus(p(ack(0, x)), y), s(y), s(z)) | | quot(0, s(y), z) | → | z |
| ack(0, x) | → | s(x) | | ack(0, x) | → | plus(x, s(0)) |
| ack(s(x), 0) | → | ack(x, s(0)) | | ack(s(x), s(y)) | → | ack(x, ack(s(x), y)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, minus, s, ack, p, div, true, false, zero, quot
Strategy
Polynomial Interpretation
- 0: 0
- ack(x,y): 0
- ack#(x,y): y
- div(x,y): 0
- false: 0
- minus(x,y): 0
- p(x): 0
- plus(x,y): 0
- quot(x,y,z): 0
- s(x): x + 2
- true: 0
- zero(x): 0
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| ack#(s(x), s(y)) | → | ack#(s(x), y) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| plus#(s(x), y) | → | plus#(y, x) | | plus#(s(x), y) | → | plus#(x, s(y)) |
Rewrite Rules
| minus(minus(x, y), z) | → | minus(x, plus(y, z)) | | minus(0, y) | → | 0 |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| plus(0, y) | → | y | | plus(s(x), y) | → | plus(x, s(y)) |
| plus(s(x), y) | → | s(plus(y, x)) | | zero(s(x)) | → | false |
| zero(0) | → | true | | p(s(x)) | → | x |
| p(0) | → | 0 | | div(x, y) | → | quot(x, y, 0) |
| quot(s(x), s(y), z) | → | quot(minus(p(ack(0, x)), y), s(y), s(z)) | | quot(0, s(y), z) | → | z |
| ack(0, x) | → | s(x) | | ack(0, x) | → | plus(x, s(0)) |
| ack(s(x), 0) | → | ack(x, s(0)) | | ack(s(x), s(y)) | → | ack(x, ack(s(x), y)) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, ack, p, div, false, true, zero, quot
Strategy
Polynomial Interpretation
- 0: 0
- ack(x,y): 0
- div(x,y): 0
- false: 0
- minus(x,y): 0
- p(x): 0
- plus(x,y): 0
- plus#(x,y): y + x
- quot(x,y,z): 0
- s(x): x + 1
- true: 0
- zero(x): 0
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| plus#(s(x), y) | → | plus#(y, x) |
Problem 7: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| plus#(s(x), y) | → | plus#(x, s(y)) |
Rewrite Rules
| minus(minus(x, y), z) | → | minus(x, plus(y, z)) | | minus(0, y) | → | 0 |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| plus(0, y) | → | y | | plus(s(x), y) | → | plus(x, s(y)) |
| plus(s(x), y) | → | s(plus(y, x)) | | zero(s(x)) | → | false |
| zero(0) | → | true | | p(s(x)) | → | x |
| p(0) | → | 0 | | div(x, y) | → | quot(x, y, 0) |
| quot(s(x), s(y), z) | → | quot(minus(p(ack(0, x)), y), s(y), s(z)) | | quot(0, s(y), z) | → | z |
| ack(0, x) | → | s(x) | | ack(0, x) | → | plus(x, s(0)) |
| ack(s(x), 0) | → | ack(x, s(0)) | | ack(s(x), s(y)) | → | ack(x, ack(s(x), y)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, minus, s, ack, p, div, true, false, zero, quot
Strategy
Polynomial Interpretation
- 0: 0
- ack(x,y): 0
- div(x,y): 0
- false: 0
- minus(x,y): 0
- p(x): 0
- plus(x,y): 0
- plus#(x,y): 2x
- quot(x,y,z): 0
- s(x): 2x + 1
- true: 0
- zero(x): 0
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| plus#(s(x), y) | → | plus#(x, s(y)) |