YES
The TRS could be proven terminating. The proof took 708 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (11ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (320ms).
| | – Problem 4 was processed with processor PolynomialLinearRange4iUR (147ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| g#(s(x), s(y)) | → | if#(f(x), s(x), s(y)) | | g#(x, c(y)) | → | g#(x, g(s(c(y)), y)) |
| f#(s(x)) | → | f#(x) | | g#(s(x), s(y)) | → | f#(x) |
| g#(x, c(y)) | → | g#(s(c(y)), y) |
Rewrite Rules
| f(0) | → | true | | f(1) | → | false |
| f(s(x)) | → | f(x) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | g(s(x), s(y)) | → | if(f(x), s(x), s(y)) |
| g(x, c(y)) | → | g(x, g(s(c(y)), y)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 1, 0, s, c, if, true, false
Strategy
The following SCCs where found
| g#(x, c(y)) → g#(x, g(s(c(y)), y)) | g#(x, c(y)) → g#(s(c(y)), y) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| g#(x, c(y)) | → | g#(x, g(s(c(y)), y)) | | g#(x, c(y)) | → | g#(s(c(y)), y) |
Rewrite Rules
| f(0) | → | true | | f(1) | → | false |
| f(s(x)) | → | f(x) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | g(s(x), s(y)) | → | if(f(x), s(x), s(y)) |
| g(x, c(y)) | → | g(x, g(s(c(y)), y)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 1, 0, s, c, if, true, false
Strategy
Polynomial Interpretation
- 0: 3
- 1: 0
- c(x): x + 3
- f(x): 0
- false: 1
- g(x,y): 3
- g#(x,y): y + x
- if(x,y,z): z + 2y
- s(x): 1
- true: 0
Improved Usable rules
| g(x, c(y)) | → | g(x, g(s(c(y)), y)) | | if(false, x, y) | → | y |
| g(s(x), s(y)) | → | if(f(x), s(x), s(y)) | | if(true, x, y) | → | x |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| g#(x, c(y)) | → | g#(s(c(y)), y) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| g#(x, c(y)) | → | g#(x, g(s(c(y)), y)) |
Rewrite Rules
| f(0) | → | true | | f(1) | → | false |
| f(s(x)) | → | f(x) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | g(s(x), s(y)) | → | if(f(x), s(x), s(y)) |
| g(x, c(y)) | → | g(x, g(s(c(y)), y)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 1, 0, s, c, if, false, true
Strategy
Polynomial Interpretation
- 0: 0
- 1: 3
- c(x): 2x + 1
- f(x): 0
- false: 3
- g(x,y): 2y
- g#(x,y): y + x + 1
- if(x,y,z): z + y
- s(x): 1
- true: 0
Improved Usable rules
| g(x, c(y)) | → | g(x, g(s(c(y)), y)) | | if(false, x, y) | → | y |
| g(s(x), s(y)) | → | if(f(x), s(x), s(y)) | | if(true, x, y) | → | x |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| g#(x, c(y)) | → | g#(x, g(s(c(y)), y)) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(0) | → | true | | f(1) | → | false |
| f(s(x)) | → | f(x) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | g(s(x), s(y)) | → | if(f(x), s(x), s(y)) |
| g(x, c(y)) | → | g(x, g(s(c(y)), y)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 1, 0, s, c, if, true, false
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: