MAYBE
The TRS could not be proven terminating. The proof attempt took 265 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (0ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (56ms), DependencyGraph (0ms), PolynomialLinearRange8NegiUR (35ms), DependencyGraph (1ms), ReductionPairSAT (25ms), DependencyGraph (1ms), SizeChangePrinciple (4ms)].
The following open problems remain:
Open Dependency Pair Problem 4
Dependency Pairs
Rewrite Rules
| and(true, X) | → | X | | and(false, Y) | → | false |
| if(true, X, Y) | → | X | | if(false, X, Y) | → | Y |
| add(0, X) | → | X | | add(s(X), Y) | → | s(add(X, Y)) |
| first(0, X) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
| from(X) | → | cons(X, from(s(X))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, if, false, true, from, add, first, cons, nil, and
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| add#(s(X), Y) | → | add#(X, Y) | | first#(s(X), cons(Y, Z)) | → | first#(X, Z) |
| from#(X) | → | from#(s(X)) |
Rewrite Rules
| and(true, X) | → | X | | and(false, Y) | → | false |
| if(true, X, Y) | → | X | | if(false, X, Y) | → | Y |
| add(0, X) | → | X | | add(s(X), Y) | → | s(add(X, Y)) |
| first(0, X) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
| from(X) | → | cons(X, from(s(X))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, if, true, false, from, add, first, and, nil, cons
Strategy
The following SCCs where found
| add#(s(X), Y) → add#(X, Y) |
| first#(s(X), cons(Y, Z)) → first#(X, Z) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| first#(s(X), cons(Y, Z)) | → | first#(X, Z) |
Rewrite Rules
| and(true, X) | → | X | | and(false, Y) | → | false |
| if(true, X, Y) | → | X | | if(false, X, Y) | → | Y |
| add(0, X) | → | X | | add(s(X), Y) | → | s(add(X, Y)) |
| first(0, X) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
| from(X) | → | cons(X, from(s(X))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, if, true, false, from, add, first, and, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| first#(s(X), cons(Y, Z)) | → | first#(X, Z) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| add#(s(X), Y) | → | add#(X, Y) |
Rewrite Rules
| and(true, X) | → | X | | and(false, Y) | → | false |
| if(true, X, Y) | → | X | | if(false, X, Y) | → | Y |
| add(0, X) | → | X | | add(s(X), Y) | → | s(add(X, Y)) |
| first(0, X) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
| from(X) | → | cons(X, from(s(X))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, if, true, false, from, add, first, and, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| add#(s(X), Y) | → | add#(X, Y) |