MAYBE
The TRS could not be proven terminating. The proof attempt took 397 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (0ms).
| – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (42ms), PolynomialLinearRange4iUR (65ms), DependencyGraph (0ms), PolynomialLinearRange8NegiUR (81ms), DependencyGraph (1ms), ReductionPairSAT (25ms), DependencyGraph (3ms), SizeChangePrinciple (4ms)].
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor SubtermCriterion (0ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
Rewrite Rules
| fact(X) | → | if(zero(X), s(0), prod(X, fact(p(X)))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | prod(0, X) | → | 0 |
| prod(s(X), Y) | → | add(Y, prod(X, Y)) | | if(true, X, Y) | → | X |
| if(false, X, Y) | → | Y | | zero(0) | → | true |
| zero(s(X)) | → | false | | p(s(X)) | → | X |
Original Signature
Termination of terms over the following signature is verified: fact, 0, s, if, p, false, true, add, zero, prod
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| add#(s(X), Y) | → | add#(X, Y) | | prod#(s(X), Y) | → | prod#(X, Y) |
| prod#(s(X), Y) | → | add#(Y, prod(X, Y)) | | fact#(X) | → | zero#(X) |
| fact#(X) | → | fact#(p(X)) | | fact#(X) | → | if#(zero(X), s(0), prod(X, fact(p(X)))) |
| fact#(X) | → | prod#(X, fact(p(X))) | | fact#(X) | → | p#(X) |
Rewrite Rules
| fact(X) | → | if(zero(X), s(0), prod(X, fact(p(X)))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | prod(0, X) | → | 0 |
| prod(s(X), Y) | → | add(Y, prod(X, Y)) | | if(true, X, Y) | → | X |
| if(false, X, Y) | → | Y | | zero(0) | → | true |
| zero(s(X)) | → | false | | p(s(X)) | → | X |
Original Signature
Termination of terms over the following signature is verified: 0, fact, s, if, p, true, false, zero, add, prod
Strategy
The following SCCs where found
| add#(s(X), Y) → add#(X, Y) |
| prod#(s(X), Y) → prod#(X, Y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| prod#(s(X), Y) | → | prod#(X, Y) |
Rewrite Rules
| fact(X) | → | if(zero(X), s(0), prod(X, fact(p(X)))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | prod(0, X) | → | 0 |
| prod(s(X), Y) | → | add(Y, prod(X, Y)) | | if(true, X, Y) | → | X |
| if(false, X, Y) | → | Y | | zero(0) | → | true |
| zero(s(X)) | → | false | | p(s(X)) | → | X |
Original Signature
Termination of terms over the following signature is verified: 0, fact, s, if, p, true, false, zero, add, prod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| prod#(s(X), Y) | → | prod#(X, Y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| add#(s(X), Y) | → | add#(X, Y) |
Rewrite Rules
| fact(X) | → | if(zero(X), s(0), prod(X, fact(p(X)))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | prod(0, X) | → | 0 |
| prod(s(X), Y) | → | add(Y, prod(X, Y)) | | if(true, X, Y) | → | X |
| if(false, X, Y) | → | Y | | zero(0) | → | true |
| zero(s(X)) | → | false | | p(s(X)) | → | X |
Original Signature
Termination of terms over the following signature is verified: 0, fact, s, if, p, true, false, zero, add, prod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| add#(s(X), Y) | → | add#(X, Y) |