YES
The TRS could be proven terminating. The proof took 221 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (19ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor PolynomialLinearRange4iUR (156ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| minus#(s(X), s(Y)) | → | minus#(X, Y) | | div#(s(X), s(Y)) | → | if#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0) |
| div#(s(X), s(Y)) | → | div#(minus(X, Y), s(Y)) | | div#(s(X), s(Y)) | → | geq#(X, Y) |
| div#(s(X), s(Y)) | → | minus#(X, Y) | | geq#(s(X), s(Y)) | → | geq#(X, Y) |
Rewrite Rules
| minus(0, Y) | → | 0 | | minus(s(X), s(Y)) | → | minus(X, Y) |
| geq(X, 0) | → | true | | geq(0, s(Y)) | → | false |
| geq(s(X), s(Y)) | → | geq(X, Y) | | div(0, s(Y)) | → | 0 |
| div(s(X), s(Y)) | → | if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0) | | if(true, X, Y) | → | X |
| if(false, X, Y) | → | Y |
Original Signature
Termination of terms over the following signature is verified: geq, minus, 0, s, if, div, true, false
Strategy
The following SCCs where found
| minus#(s(X), s(Y)) → minus#(X, Y) |
| div#(s(X), s(Y)) → div#(minus(X, Y), s(Y)) |
| geq#(s(X), s(Y)) → geq#(X, Y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(s(X), s(Y)) | → | minus#(X, Y) |
Rewrite Rules
| minus(0, Y) | → | 0 | | minus(s(X), s(Y)) | → | minus(X, Y) |
| geq(X, 0) | → | true | | geq(0, s(Y)) | → | false |
| geq(s(X), s(Y)) | → | geq(X, Y) | | div(0, s(Y)) | → | 0 |
| div(s(X), s(Y)) | → | if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0) | | if(true, X, Y) | → | X |
| if(false, X, Y) | → | Y |
Original Signature
Termination of terms over the following signature is verified: geq, minus, 0, s, if, div, true, false
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(X), s(Y)) | → | minus#(X, Y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| geq#(s(X), s(Y)) | → | geq#(X, Y) |
Rewrite Rules
| minus(0, Y) | → | 0 | | minus(s(X), s(Y)) | → | minus(X, Y) |
| geq(X, 0) | → | true | | geq(0, s(Y)) | → | false |
| geq(s(X), s(Y)) | → | geq(X, Y) | | div(0, s(Y)) | → | 0 |
| div(s(X), s(Y)) | → | if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0) | | if(true, X, Y) | → | X |
| if(false, X, Y) | → | Y |
Original Signature
Termination of terms over the following signature is verified: geq, minus, 0, s, if, div, true, false
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| geq#(s(X), s(Y)) | → | geq#(X, Y) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| div#(s(X), s(Y)) | → | div#(minus(X, Y), s(Y)) |
Rewrite Rules
| minus(0, Y) | → | 0 | | minus(s(X), s(Y)) | → | minus(X, Y) |
| geq(X, 0) | → | true | | geq(0, s(Y)) | → | false |
| geq(s(X), s(Y)) | → | geq(X, Y) | | div(0, s(Y)) | → | 0 |
| div(s(X), s(Y)) | → | if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0) | | if(true, X, Y) | → | X |
| if(false, X, Y) | → | Y |
Original Signature
Termination of terms over the following signature is verified: geq, minus, 0, s, if, div, true, false
Strategy
Polynomial Interpretation
- 0: 0
- div(x,y): 0
- div#(x,y): x + 1
- false: 0
- geq(x,y): 0
- if(x,y,z): 0
- minus(x,y): 0
- s(x): 2
- true: 0
Improved Usable rules
| minus(s(X), s(Y)) | → | minus(X, Y) | | minus(0, Y) | → | 0 |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| div#(s(X), s(Y)) | → | div#(minus(X, Y), s(Y)) |