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 (118ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (4ms), PolynomialLinearRange4iUR (1388ms), DependencyGraph (5ms), PolynomialLinearRange8NegiUR (6127ms), DependencyGraph (3ms), ReductionPairSAT (1476ms), DependencyGraph (7ms), SizeChangePrinciple (490ms), ForwardNarrowing (3ms), BackwardInstantiation (2ms), ForwardInstantiation (2ms), Propagation (0ms)].
| – Problem 4 was processed with processor SubtermCriterion (2ms).
The following open problems remain:
Open Dependency Pair Problem 3
Dependency Pairs
| plus#(id(x), s(y)) | → | plus#(x, if(gt(s(y), y), y, s(y))) | | plus#(s(x), s(y)) | → | plus#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))) |
| plus#(s(x), x) | → | plus#(if(gt(x, x), id(x), id(x)), s(x)) |
Rewrite Rules
| times(x, 0) | → | 0 | | times(x, s(y)) | → | plus(times(x, y), x) |
| plus(s(x), s(y)) | → | s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) | | plus(s(x), x) | → | plus(if(gt(x, x), id(x), id(x)), s(x)) |
| plus(zero, y) | → | y | | plus(id(x), s(y)) | → | s(plus(x, if(gt(s(y), y), y, s(y)))) |
| id(x) | → | x | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | not(x) | → | if(x, false, true) |
| gt(s(x), zero) | → | true | | gt(zero, y) | → | false |
| gt(s(x), s(y)) | → | gt(x, y) |
Original Signature
Termination of terms over the following signature is verified: not, id, plus, 0, s, times, if, false, true, gt, zero
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| plus#(s(x), x) | → | id#(x) | | plus#(id(x), s(y)) | → | plus#(x, if(gt(s(y), y), y, s(y))) |
| plus#(id(x), s(y)) | → | gt#(s(y), y) | | plus#(s(x), s(y)) | → | gt#(x, y) |
| plus#(s(x), s(y)) | → | if#(gt(x, y), x, y) | | plus#(s(x), x) | → | plus#(if(gt(x, x), id(x), id(x)), s(x)) |
| plus#(s(x), x) | → | gt#(x, x) | | plus#(id(x), s(y)) | → | if#(gt(s(y), y), y, s(y)) |
| times#(x, s(y)) | → | times#(x, y) | | plus#(s(x), s(y)) | → | id#(x) |
| plus#(s(x), x) | → | if#(gt(x, x), id(x), id(x)) | | plus#(s(x), s(y)) | → | plus#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))) |
| times#(x, s(y)) | → | plus#(times(x, y), x) | | plus#(s(x), s(y)) | → | if#(not(gt(x, y)), id(x), id(y)) |
| gt#(s(x), s(y)) | → | gt#(x, y) | | not#(x) | → | if#(x, false, true) |
| plus#(s(x), s(y)) | → | id#(y) | | plus#(s(x), s(y)) | → | not#(gt(x, y)) |
Rewrite Rules
| times(x, 0) | → | 0 | | times(x, s(y)) | → | plus(times(x, y), x) |
| plus(s(x), s(y)) | → | s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) | | plus(s(x), x) | → | plus(if(gt(x, x), id(x), id(x)), s(x)) |
| plus(zero, y) | → | y | | plus(id(x), s(y)) | → | s(plus(x, if(gt(s(y), y), y, s(y)))) |
| id(x) | → | x | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | not(x) | → | if(x, false, true) |
| gt(s(x), zero) | → | true | | gt(zero, y) | → | false |
| gt(s(x), s(y)) | → | gt(x, y) |
Original Signature
Termination of terms over the following signature is verified: id, not, plus, 0, s, times, if, true, false, gt, zero
Strategy
The following SCCs where found
| plus#(id(x), s(y)) → plus#(x, if(gt(s(y), y), y, s(y))) | plus#(s(x), s(y)) → plus#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))) |
| plus#(s(x), x) → plus#(if(gt(x, x), id(x), id(x)), s(x)) |
| gt#(s(x), s(y)) → gt#(x, y) |
| times#(x, s(y)) → times#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| times#(x, s(y)) | → | times#(x, y) |
Rewrite Rules
| times(x, 0) | → | 0 | | times(x, s(y)) | → | plus(times(x, y), x) |
| plus(s(x), s(y)) | → | s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) | | plus(s(x), x) | → | plus(if(gt(x, x), id(x), id(x)), s(x)) |
| plus(zero, y) | → | y | | plus(id(x), s(y)) | → | s(plus(x, if(gt(s(y), y), y, s(y)))) |
| id(x) | → | x | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | not(x) | → | if(x, false, true) |
| gt(s(x), zero) | → | true | | gt(zero, y) | → | false |
| gt(s(x), s(y)) | → | gt(x, y) |
Original Signature
Termination of terms over the following signature is verified: id, not, plus, 0, s, times, if, true, false, gt, zero
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| times#(x, s(y)) | → | times#(x, y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| gt#(s(x), s(y)) | → | gt#(x, y) |
Rewrite Rules
| times(x, 0) | → | 0 | | times(x, s(y)) | → | plus(times(x, y), x) |
| plus(s(x), s(y)) | → | s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) | | plus(s(x), x) | → | plus(if(gt(x, x), id(x), id(x)), s(x)) |
| plus(zero, y) | → | y | | plus(id(x), s(y)) | → | s(plus(x, if(gt(s(y), y), y, s(y)))) |
| id(x) | → | x | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | not(x) | → | if(x, false, true) |
| gt(s(x), zero) | → | true | | gt(zero, y) | → | false |
| gt(s(x), s(y)) | → | gt(x, y) |
Original Signature
Termination of terms over the following signature is verified: id, not, plus, 0, s, times, if, true, false, gt, zero
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| gt#(s(x), s(y)) | → | gt#(x, y) |