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 (44ms).
| – Problem 2 was processed with processor SubtermCriterion (2ms).
| – Problem 3 was processed with processor SubtermCriterion (2ms).
| – Problem 4 was processed with processor SubtermCriterion (0ms).
| – Problem 5 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (795ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (14723ms), DependencyGraph (1ms), ReductionPairSAT (1531ms), DependencyGraph (1ms), SizeChangePrinciple (819ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (0ms), Propagation (1ms)].
The following open problems remain:
Open Dependency Pair Problem 5
Dependency Pairs
| f#(true, x, y) | → | f#(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y)) |
Rewrite Rules
| f(true, x, y) | → | f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y)) | | gt(0, v) | → | false |
| gt(s(u), 0) | → | true | | gt(s(u), s(v)) | → | gt(u, v) |
| and(x, true) | → | x | | and(x, false) | → | false |
| plus(n, 0) | → | n | | plus(n, s(m)) | → | s(plus(n, m)) |
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
Original Signature
Termination of terms over the following signature is verified: f, plus, 0, s, false, true, gt, double, and
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(true, x, y) | → | gt#(x, y) | | f#(true, x, y) | → | double#(y) |
| f#(true, x, y) | → | and#(gt(x, y), gt(y, s(s(0)))) | | f#(true, x, y) | → | f#(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y)) |
| gt#(s(u), s(v)) | → | gt#(u, v) | | double#(s(x)) | → | double#(x) |
| f#(true, x, y) | → | plus#(s(0), x) | | plus#(n, s(m)) | → | plus#(n, m) |
| f#(true, x, y) | → | gt#(y, s(s(0))) |
Rewrite Rules
| f(true, x, y) | → | f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y)) | | gt(0, v) | → | false |
| gt(s(u), 0) | → | true | | gt(s(u), s(v)) | → | gt(u, v) |
| and(x, true) | → | x | | and(x, false) | → | false |
| plus(n, 0) | → | n | | plus(n, s(m)) | → | s(plus(n, m)) |
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
Original Signature
Termination of terms over the following signature is verified: f, plus, 0, s, true, false, gt, double, and
Strategy
The following SCCs where found
| f#(true, x, y) → f#(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y)) |
| double#(s(x)) → double#(x) |
| gt#(s(u), s(v)) → gt#(u, v) |
| plus#(n, s(m)) → plus#(n, m) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| gt#(s(u), s(v)) | → | gt#(u, v) |
Rewrite Rules
| f(true, x, y) | → | f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y)) | | gt(0, v) | → | false |
| gt(s(u), 0) | → | true | | gt(s(u), s(v)) | → | gt(u, v) |
| and(x, true) | → | x | | and(x, false) | → | false |
| plus(n, 0) | → | n | | plus(n, s(m)) | → | s(plus(n, m)) |
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
Original Signature
Termination of terms over the following signature is verified: f, plus, 0, s, true, false, gt, double, and
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| gt#(s(u), s(v)) | → | gt#(u, v) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| double#(s(x)) | → | double#(x) |
Rewrite Rules
| f(true, x, y) | → | f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y)) | | gt(0, v) | → | false |
| gt(s(u), 0) | → | true | | gt(s(u), s(v)) | → | gt(u, v) |
| and(x, true) | → | x | | and(x, false) | → | false |
| plus(n, 0) | → | n | | plus(n, s(m)) | → | s(plus(n, m)) |
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
Original Signature
Termination of terms over the following signature is verified: f, plus, 0, s, true, false, gt, double, and
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| double#(s(x)) | → | double#(x) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(n, s(m)) | → | plus#(n, m) |
Rewrite Rules
| f(true, x, y) | → | f(and(gt(x, y), gt(y, s(s(0)))), plus(s(0), x), double(y)) | | gt(0, v) | → | false |
| gt(s(u), 0) | → | true | | gt(s(u), s(v)) | → | gt(u, v) |
| and(x, true) | → | x | | and(x, false) | → | false |
| plus(n, 0) | → | n | | plus(n, s(m)) | → | s(plus(n, m)) |
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
Original Signature
Termination of terms over the following signature is verified: f, plus, 0, s, true, false, gt, double, and
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| plus#(n, s(m)) | → | plus#(n, m) |