YES
The TRS could be proven terminating. The proof took 424 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (64ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| | – Problem 5 was processed with processor PolynomialLinearRange4iUR (211ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(s(x), s(y), z, u) | → | le#(x, y) | | le#(s(x), s(y)) | → | le#(x, y) |
| f#(s(x), s(y), z, u) | → | f#(s(x), minus(y, x), z, u) | | minus#(s(x), s(y)) | → | minus#(x, y) |
| f#(s(x), 0, z, u) | → | f#(x, u, minus(z, s(x)), u) | | f#(s(x), s(y), z, u) | → | minus#(y, x) |
| f#(s(x), s(y), z, u) | → | f#(x, u, z, u) | | f#(s(x), s(y), z, u) | → | if#(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) |
| f#(s(x), 0, z, u) | → | minus#(z, s(x)) | | perfectp#(s(x)) | → | f#(x, s(0), s(x), s(x)) |
Rewrite Rules
| minus(0, y) | → | 0 | | minus(s(x), 0) | → | s(x) |
| minus(s(x), s(y)) | → | minus(x, y) | | le(0, y) | → | true |
| le(s(x), 0) | → | false | | le(s(x), s(y)) | → | le(x, y) |
| if(true, x, y) | → | x | | if(false, x, y) | → | y |
| perfectp(0) | → | false | | perfectp(s(x)) | → | f(x, s(0), s(x), s(x)) |
| f(0, y, 0, u) | → | true | | f(0, y, s(z), u) | → | false |
| f(s(x), 0, z, u) | → | f(x, u, minus(z, s(x)), u) | | f(s(x), s(y), z, u) | → | if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) |
Original Signature
Termination of terms over the following signature is verified: f, minus, 0, s, le, if, perfectp, true, false
Strategy
The following SCCs where found
| le#(s(x), s(y)) → le#(x, y) |
| minus#(s(x), s(y)) → minus#(x, y) |
| f#(s(x), s(y), z, u) → f#(s(x), minus(y, x), z, u) | f#(s(x), s(y), z, u) → f#(x, u, z, u) |
| f#(s(x), 0, z, u) → f#(x, u, minus(z, s(x)), u) |
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), 0) | → | s(x) |
| minus(s(x), s(y)) | → | minus(x, y) | | le(0, y) | → | true |
| le(s(x), 0) | → | false | | le(s(x), s(y)) | → | le(x, y) |
| if(true, x, y) | → | x | | if(false, x, y) | → | y |
| perfectp(0) | → | false | | perfectp(s(x)) | → | f(x, s(0), s(x), s(x)) |
| f(0, y, 0, u) | → | true | | f(0, y, s(z), u) | → | false |
| f(s(x), 0, z, u) | → | f(x, u, minus(z, s(x)), u) | | f(s(x), s(y), z, u) | → | if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) |
Original Signature
Termination of terms over the following signature is verified: f, minus, 0, s, le, if, perfectp, 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
| f#(s(x), s(y), z, u) | → | f#(s(x), minus(y, x), z, u) | | f#(s(x), s(y), z, u) | → | f#(x, u, z, u) |
| f#(s(x), 0, z, u) | → | f#(x, u, minus(z, s(x)), u) |
Rewrite Rules
| minus(0, y) | → | 0 | | minus(s(x), 0) | → | s(x) |
| minus(s(x), s(y)) | → | minus(x, y) | | le(0, y) | → | true |
| le(s(x), 0) | → | false | | le(s(x), s(y)) | → | le(x, y) |
| if(true, x, y) | → | x | | if(false, x, y) | → | y |
| perfectp(0) | → | false | | perfectp(s(x)) | → | f(x, s(0), s(x), s(x)) |
| f(0, y, 0, u) | → | true | | f(0, y, s(z), u) | → | false |
| f(s(x), 0, z, u) | → | f(x, u, minus(z, s(x)), u) | | f(s(x), s(y), z, u) | → | if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) |
Original Signature
Termination of terms over the following signature is verified: f, minus, 0, s, le, if, perfectp, true, false
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(s(x), s(y), z, u) | → | f#(x, u, z, u) | | f#(s(x), 0, z, u) | → | f#(x, u, minus(z, s(x)), u) |
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(s(x), s(y), z, u) | → | f#(s(x), minus(y, x), z, u) |
Rewrite Rules
| minus(0, y) | → | 0 | | minus(s(x), 0) | → | s(x) |
| minus(s(x), s(y)) | → | minus(x, y) | | le(0, y) | → | true |
| le(s(x), 0) | → | false | | le(s(x), s(y)) | → | le(x, y) |
| if(true, x, y) | → | x | | if(false, x, y) | → | y |
| perfectp(0) | → | false | | perfectp(s(x)) | → | f(x, s(0), s(x), s(x)) |
| f(0, y, 0, u) | → | true | | f(0, y, s(z), u) | → | false |
| f(s(x), 0, z, u) | → | f(x, u, minus(z, s(x)), u) | | f(s(x), s(y), z, u) | → | if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) |
Original Signature
Termination of terms over the following signature is verified: f, 0, minus, le, s, if, perfectp, false, true
Strategy
Polynomial Interpretation
- 0: 1
- f(x1,x2,x3,x4): 0
- f#(x1,x2,x3,x4): x4 + x3 + 2x2
- false: 0
- if(x,y,z): 0
- le(x,y): 0
- minus(x,y): x
- perfectp(x): 0
- s(x): x + 1
- true: 0
Improved Usable rules
| minus(s(x), s(y)) | → | minus(x, y) | | minus(0, y) | → | 0 |
| minus(s(x), 0) | → | s(x) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| f#(s(x), s(y), z, u) | → | f#(s(x), minus(y, x), z, u) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| le#(s(x), s(y)) | → | le#(x, y) |
Rewrite Rules
| minus(0, y) | → | 0 | | minus(s(x), 0) | → | s(x) |
| minus(s(x), s(y)) | → | minus(x, y) | | le(0, y) | → | true |
| le(s(x), 0) | → | false | | le(s(x), s(y)) | → | le(x, y) |
| if(true, x, y) | → | x | | if(false, x, y) | → | y |
| perfectp(0) | → | false | | perfectp(s(x)) | → | f(x, s(0), s(x), s(x)) |
| f(0, y, 0, u) | → | true | | f(0, y, s(z), u) | → | false |
| f(s(x), 0, z, u) | → | f(x, u, minus(z, s(x)), u) | | f(s(x), s(y), z, u) | → | if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) |
Original Signature
Termination of terms over the following signature is verified: f, minus, 0, s, le, if, perfectp, true, false
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| le#(s(x), s(y)) | → | le#(x, y) |