YES
The TRS could be proven terminating. The proof took 988 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (123ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (200ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (144ms).
| – Problem 4 was processed with processor PolynomialLinearRange4iUR (350ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| minus#(n__s(X), n__s(Y)) | → | minus#(activate(X), activate(Y)) | | minus#(n__s(X), n__s(Y)) | → | activate#(Y) |
| geq#(n__s(X), n__s(Y)) | → | activate#(X) | | div#(s(X), n__s(Y)) | → | if#(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0) |
| activate#(n__s(X)) | → | s#(X) | | div#(s(X), n__s(Y)) | → | activate#(Y) |
| div#(s(X), n__s(Y)) | → | div#(minus(X, activate(Y)), n__s(activate(Y))) | | minus#(n__0, Y) | → | 0# |
| geq#(n__s(X), n__s(Y)) | → | geq#(activate(X), activate(Y)) | | if#(false, X, Y) | → | activate#(Y) |
| if#(true, X, Y) | → | activate#(X) | | div#(s(X), n__s(Y)) | → | minus#(X, activate(Y)) |
| activate#(n__0) | → | 0# | | div#(0, n__s(Y)) | → | 0# |
| div#(s(X), n__s(Y)) | → | geq#(X, activate(Y)) | | geq#(n__s(X), n__s(Y)) | → | activate#(Y) |
| minus#(n__s(X), n__s(Y)) | → | activate#(X) |
Rewrite Rules
| minus(n__0, Y) | → | 0 | | minus(n__s(X), n__s(Y)) | → | minus(activate(X), activate(Y)) |
| geq(X, n__0) | → | true | | geq(n__0, n__s(Y)) | → | false |
| geq(n__s(X), n__s(Y)) | → | geq(activate(X), activate(Y)) | | div(0, n__s(Y)) | → | 0 |
| div(s(X), n__s(Y)) | → | if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0) | | if(true, X, Y) | → | activate(X) |
| if(false, X, Y) | → | activate(Y) | | 0 | → | n__0 |
| s(X) | → | n__s(X) | | activate(n__0) | → | 0 |
| activate(n__s(X)) | → | s(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, n__s, geq, 0, minus, n__0, s, if, div, true, false
Strategy
The following SCCs where found
| div#(s(X), n__s(Y)) → div#(minus(X, activate(Y)), n__s(activate(Y))) |
| geq#(n__s(X), n__s(Y)) → geq#(activate(X), activate(Y)) |
| minus#(n__s(X), n__s(Y)) → minus#(activate(X), activate(Y)) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| geq#(n__s(X), n__s(Y)) | → | geq#(activate(X), activate(Y)) |
Rewrite Rules
| minus(n__0, Y) | → | 0 | | minus(n__s(X), n__s(Y)) | → | minus(activate(X), activate(Y)) |
| geq(X, n__0) | → | true | | geq(n__0, n__s(Y)) | → | false |
| geq(n__s(X), n__s(Y)) | → | geq(activate(X), activate(Y)) | | div(0, n__s(Y)) | → | 0 |
| div(s(X), n__s(Y)) | → | if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0) | | if(true, X, Y) | → | activate(X) |
| if(false, X, Y) | → | activate(Y) | | 0 | → | n__0 |
| s(X) | → | n__s(X) | | activate(n__0) | → | 0 |
| activate(n__s(X)) | → | s(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, n__s, geq, 0, minus, n__0, s, if, div, true, false
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): x + 1
- div(x,y): 0
- false: 0
- geq(x,y): 0
- geq#(x,y): y
- if(x,y,z): 0
- minus(x,y): 0
- n__0: 0
- n__s(x): x + 2
- s(x): x + 3
- true: 0
Improved Usable rules
| s(X) | → | n__s(X) | | activate(X) | → | X |
| activate(n__s(X)) | → | s(X) | | activate(n__0) | → | 0 |
| 0 | → | n__0 |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| geq#(n__s(X), n__s(Y)) | → | geq#(activate(X), activate(Y)) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| minus#(n__s(X), n__s(Y)) | → | minus#(activate(X), activate(Y)) |
Rewrite Rules
| minus(n__0, Y) | → | 0 | | minus(n__s(X), n__s(Y)) | → | minus(activate(X), activate(Y)) |
| geq(X, n__0) | → | true | | geq(n__0, n__s(Y)) | → | false |
| geq(n__s(X), n__s(Y)) | → | geq(activate(X), activate(Y)) | | div(0, n__s(Y)) | → | 0 |
| div(s(X), n__s(Y)) | → | if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0) | | if(true, X, Y) | → | activate(X) |
| if(false, X, Y) | → | activate(Y) | | 0 | → | n__0 |
| s(X) | → | n__s(X) | | activate(n__0) | → | 0 |
| activate(n__s(X)) | → | s(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, n__s, geq, 0, minus, n__0, s, if, div, true, false
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): 2x
- div(x,y): 0
- false: 0
- geq(x,y): 0
- if(x,y,z): 0
- minus(x,y): 0
- minus#(x,y): y + 1
- n__0: 0
- n__s(x): 2x + 1
- s(x): 3x + 1
- true: 0
Improved Usable rules
| s(X) | → | n__s(X) | | activate(X) | → | X |
| activate(n__s(X)) | → | s(X) | | activate(n__0) | → | 0 |
| 0 | → | n__0 |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| minus#(n__s(X), n__s(Y)) | → | minus#(activate(X), activate(Y)) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| div#(s(X), n__s(Y)) | → | div#(minus(X, activate(Y)), n__s(activate(Y))) |
Rewrite Rules
| minus(n__0, Y) | → | 0 | | minus(n__s(X), n__s(Y)) | → | minus(activate(X), activate(Y)) |
| geq(X, n__0) | → | true | | geq(n__0, n__s(Y)) | → | false |
| geq(n__s(X), n__s(Y)) | → | geq(activate(X), activate(Y)) | | div(0, n__s(Y)) | → | 0 |
| div(s(X), n__s(Y)) | → | if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0) | | if(true, X, Y) | → | activate(X) |
| if(false, X, Y) | → | activate(Y) | | 0 | → | n__0 |
| s(X) | → | n__s(X) | | activate(n__0) | → | 0 |
| activate(n__s(X)) | → | s(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, n__s, geq, 0, minus, n__0, s, if, div, true, false
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): 2
- div(x,y): 0
- div#(x,y): x + 1
- false: 0
- geq(x,y): 0
- if(x,y,z): 0
- minus(x,y): 1
- n__0: 0
- n__s(x): 0
- s(x): 2
- true: 0
Improved Usable rules
| minus(n__0, Y) | → | 0 | | minus(n__s(X), n__s(Y)) | → | minus(activate(X), activate(Y)) |
| 0 | → | n__0 |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| div#(s(X), n__s(Y)) | → | div#(minus(X, activate(Y)), n__s(activate(Y))) |