YES
The TRS could be proven terminating. The proof took 40920 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (234ms).
| – Problem 2 was processed with processor PolynomialLinearRange4 (173ms).
| | – Problem 3 was processed with processor PolynomialLinearRange4 (114ms).
| | | – Problem 4 was processed with processor DependencyGraph (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| and#(tt, X) | → | activate#(X) | | U11#(tt, N) | → | activate#(N) |
| U21#(tt, M, N) | → | s#(plus(activate(N), activate(M))) | | activate#(n__s(X)) | → | activate#(X) |
| activate#(n__plus(X1, X2)) | → | plus#(activate(X1), activate(X2)) | | plus#(N, s(M)) | → | U21#(and(isNat(M), n__isNat(N)), M, N) |
| isNat#(n__plus(V1, V2)) | → | isNat#(activate(V1)) | | U21#(tt, M, N) | → | plus#(activate(N), activate(M)) |
| plus#(N, 0) | → | isNat#(N) | | U21#(tt, M, N) | → | activate#(N) |
| isNat#(n__plus(V1, V2)) | → | activate#(V2) | | isNat#(n__plus(V1, V2)) | → | and#(isNat(activate(V1)), n__isNat(activate(V2))) |
| plus#(N, s(M)) | → | isNat#(M) | | activate#(n__plus(X1, X2)) | → | activate#(X2) |
| plus#(N, s(M)) | → | and#(isNat(M), n__isNat(N)) | | isNat#(n__s(V1)) | → | isNat#(activate(V1)) |
| plus#(N, 0) | → | U11#(isNat(N), N) | | U21#(tt, M, N) | → | activate#(M) |
| activate#(n__isNat(X)) | → | isNat#(X) | | activate#(n__0) | → | 0# |
| isNat#(n__plus(V1, V2)) | → | activate#(V1) | | activate#(n__s(X)) | → | s#(activate(X)) |
| isNat#(n__s(V1)) | → | activate#(V1) | | activate#(n__plus(X1, X2)) | → | activate#(X1) |
Rewrite Rules
| U11(tt, N) | → | activate(N) | | U21(tt, M, N) | → | s(plus(activate(N), activate(M))) |
| and(tt, X) | → | activate(X) | | isNat(n__0) | → | tt |
| isNat(n__plus(V1, V2)) | → | and(isNat(activate(V1)), n__isNat(activate(V2))) | | isNat(n__s(V1)) | → | isNat(activate(V1)) |
| plus(N, 0) | → | U11(isNat(N), N) | | plus(N, s(M)) | → | U21(and(isNat(M), n__isNat(N)), M, N) |
| 0 | → | n__0 | | plus(X1, X2) | → | n__plus(X1, X2) |
| isNat(X) | → | n__isNat(X) | | s(X) | → | n__s(X) |
| activate(n__0) | → | 0 | | activate(n__plus(X1, X2)) | → | plus(activate(X1), activate(X2)) |
| activate(n__isNat(X)) | → | isNat(X) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: plus, n__isNat, n__plus, and, activate, isNat, n__s, n__0, 0, s, tt, U11, U21
Strategy
The following SCCs where found
| and#(tt, X) → activate#(X) | U11#(tt, N) → activate#(N) |
| activate#(n__s(X)) → activate#(X) | activate#(n__plus(X1, X2)) → plus#(activate(X1), activate(X2)) |
| plus#(N, s(M)) → U21#(and(isNat(M), n__isNat(N)), M, N) | isNat#(n__plus(V1, V2)) → isNat#(activate(V1)) |
| U21#(tt, M, N) → plus#(activate(N), activate(M)) | plus#(N, 0) → isNat#(N) |
| U21#(tt, M, N) → activate#(N) | isNat#(n__plus(V1, V2)) → activate#(V2) |
| isNat#(n__plus(V1, V2)) → and#(isNat(activate(V1)), n__isNat(activate(V2))) | activate#(n__plus(X1, X2)) → activate#(X2) |
| plus#(N, s(M)) → isNat#(M) | plus#(N, s(M)) → and#(isNat(M), n__isNat(N)) |
| isNat#(n__s(V1)) → isNat#(activate(V1)) | U21#(tt, M, N) → activate#(M) |
| plus#(N, 0) → U11#(isNat(N), N) | activate#(n__isNat(X)) → isNat#(X) |
| isNat#(n__plus(V1, V2)) → activate#(V1) | isNat#(n__s(V1)) → activate#(V1) |
| activate#(n__plus(X1, X2)) → activate#(X1) |
Problem 2: PolynomialLinearRange4
Dependency Pair Problem
Dependency Pairs
| and#(tt, X) | → | activate#(X) | | U11#(tt, N) | → | activate#(N) |
| activate#(n__plus(X1, X2)) | → | plus#(activate(X1), activate(X2)) | | activate#(n__s(X)) | → | activate#(X) |
| isNat#(n__plus(V1, V2)) | → | isNat#(activate(V1)) | | plus#(N, s(M)) | → | U21#(and(isNat(M), n__isNat(N)), M, N) |
| U21#(tt, M, N) | → | plus#(activate(N), activate(M)) | | isNat#(n__plus(V1, V2)) | → | activate#(V2) |
| U21#(tt, M, N) | → | activate#(N) | | plus#(N, 0) | → | isNat#(N) |
| isNat#(n__plus(V1, V2)) | → | and#(isNat(activate(V1)), n__isNat(activate(V2))) | | plus#(N, s(M)) | → | isNat#(M) |
| activate#(n__plus(X1, X2)) | → | activate#(X2) | | plus#(N, s(M)) | → | and#(isNat(M), n__isNat(N)) |
| isNat#(n__s(V1)) | → | isNat#(activate(V1)) | | U21#(tt, M, N) | → | activate#(M) |
| plus#(N, 0) | → | U11#(isNat(N), N) | | activate#(n__isNat(X)) | → | isNat#(X) |
| isNat#(n__plus(V1, V2)) | → | activate#(V1) | | isNat#(n__s(V1)) | → | activate#(V1) |
| activate#(n__plus(X1, X2)) | → | activate#(X1) |
Rewrite Rules
| U11(tt, N) | → | activate(N) | | U21(tt, M, N) | → | s(plus(activate(N), activate(M))) |
| and(tt, X) | → | activate(X) | | isNat(n__0) | → | tt |
| isNat(n__plus(V1, V2)) | → | and(isNat(activate(V1)), n__isNat(activate(V2))) | | isNat(n__s(V1)) | → | isNat(activate(V1)) |
| plus(N, 0) | → | U11(isNat(N), N) | | plus(N, s(M)) | → | U21(and(isNat(M), n__isNat(N)), M, N) |
| 0 | → | n__0 | | plus(X1, X2) | → | n__plus(X1, X2) |
| isNat(X) | → | n__isNat(X) | | s(X) | → | n__s(X) |
| activate(n__0) | → | 0 | | activate(n__plus(X1, X2)) | → | plus(activate(X1), activate(X2)) |
| activate(n__isNat(X)) | → | isNat(X) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: plus, n__isNat, n__plus, and, activate, isNat, n__s, n__0, 0, s, tt, U11, U21
Strategy
Polynomial Interpretation
- 0: 0
- U11(x,y): y
- U11#(x,y): y
- U21(x,y,z): z + y + 2
- U21#(x,y,z): z + y + 1
- activate(x): x
- activate#(x): x
- and(x,y): y
- and#(x,y): y
- isNat(x): x
- isNat#(x): x
- n__0: 0
- n__isNat(x): x
- n__plus(x,y): y + x
- n__s(x): x + 2
- plus(x,y): y + x
- plus#(x,y): y + x
- s(x): x + 2
- tt: 0
Standard Usable rules
| plus(X1, X2) | → | n__plus(X1, X2) | | plus(N, 0) | → | U11(isNat(N), N) |
| isNat(n__plus(V1, V2)) | → | and(isNat(activate(V1)), n__isNat(activate(V2))) | | U21(tt, M, N) | → | s(plus(activate(N), activate(M))) |
| activate(n__0) | → | 0 | | U11(tt, N) | → | activate(N) |
| 0 | → | n__0 | | isNat(n__0) | → | tt |
| and(tt, X) | → | activate(X) | | s(X) | → | n__s(X) |
| activate(X) | → | X | | isNat(n__s(V1)) | → | isNat(activate(V1)) |
| activate(n__isNat(X)) | → | isNat(X) | | isNat(X) | → | n__isNat(X) |
| plus(N, s(M)) | → | U21(and(isNat(M), n__isNat(N)), M, N) | | activate(n__plus(X1, X2)) | → | plus(activate(X1), activate(X2)) |
| activate(n__s(X)) | → | s(activate(X)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| activate#(n__s(X)) | → | activate#(X) | | plus#(N, s(M)) | → | U21#(and(isNat(M), n__isNat(N)), M, N) |
| U21#(tt, M, N) | → | plus#(activate(N), activate(M)) | | U21#(tt, M, N) | → | activate#(N) |
| plus#(N, s(M)) | → | isNat#(M) | | plus#(N, s(M)) | → | and#(isNat(M), n__isNat(N)) |
| isNat#(n__s(V1)) | → | isNat#(activate(V1)) | | U21#(tt, M, N) | → | activate#(M) |
| isNat#(n__s(V1)) | → | activate#(V1) |
Problem 3: PolynomialLinearRange4
Dependency Pair Problem
Dependency Pairs
| isNat#(n__plus(V1, V2)) | → | and#(isNat(activate(V1)), n__isNat(activate(V2))) | | activate#(n__plus(X1, X2)) | → | activate#(X2) |
| and#(tt, X) | → | activate#(X) | | U11#(tt, N) | → | activate#(N) |
| activate#(n__plus(X1, X2)) | → | plus#(activate(X1), activate(X2)) | | plus#(N, 0) | → | U11#(isNat(N), N) |
| activate#(n__isNat(X)) | → | isNat#(X) | | isNat#(n__plus(V1, V2)) | → | isNat#(activate(V1)) |
| isNat#(n__plus(V1, V2)) | → | activate#(V1) | | activate#(n__plus(X1, X2)) | → | activate#(X1) |
| plus#(N, 0) | → | isNat#(N) | | isNat#(n__plus(V1, V2)) | → | activate#(V2) |
Rewrite Rules
| U11(tt, N) | → | activate(N) | | U21(tt, M, N) | → | s(plus(activate(N), activate(M))) |
| and(tt, X) | → | activate(X) | | isNat(n__0) | → | tt |
| isNat(n__plus(V1, V2)) | → | and(isNat(activate(V1)), n__isNat(activate(V2))) | | isNat(n__s(V1)) | → | isNat(activate(V1)) |
| plus(N, 0) | → | U11(isNat(N), N) | | plus(N, s(M)) | → | U21(and(isNat(M), n__isNat(N)), M, N) |
| 0 | → | n__0 | | plus(X1, X2) | → | n__plus(X1, X2) |
| isNat(X) | → | n__isNat(X) | | s(X) | → | n__s(X) |
| activate(n__0) | → | 0 | | activate(n__plus(X1, X2)) | → | plus(activate(X1), activate(X2)) |
| activate(n__isNat(X)) | → | isNat(X) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: plus, n__isNat, n__plus, and, activate, isNat, n__s, n__0, 0, s, tt, U11, U21
Strategy
Polynomial Interpretation
- 0: 1
- U11(x,y): y
- U11#(x,y): y + x
- U21(x,y,z): 3z + y + 1
- activate(x): x
- activate#(x): x + 2
- and(x,y): y + 2
- and#(x,y): y + x
- isNat(x): 2x
- isNat#(x): 2x
- n__0: 1
- n__isNat(x): 2x
- n__plus(x,y): y + 3x + 1
- n__s(x): x
- plus(x,y): y + 3x + 1
- plus#(x,y): 3x + 3
- s(x): x
- tt: 2
Standard Usable rules
| plus(X1, X2) | → | n__plus(X1, X2) | | plus(N, 0) | → | U11(isNat(N), N) |
| isNat(n__plus(V1, V2)) | → | and(isNat(activate(V1)), n__isNat(activate(V2))) | | U21(tt, M, N) | → | s(plus(activate(N), activate(M))) |
| activate(n__0) | → | 0 | | U11(tt, N) | → | activate(N) |
| 0 | → | n__0 | | isNat(n__0) | → | tt |
| and(tt, X) | → | activate(X) | | s(X) | → | n__s(X) |
| activate(X) | → | X | | isNat(n__s(V1)) | → | isNat(activate(V1)) |
| activate(n__isNat(X)) | → | isNat(X) | | isNat(X) | → | n__isNat(X) |
| plus(N, s(M)) | → | U21(and(isNat(M), n__isNat(N)), M, N) | | activate(n__plus(X1, X2)) | → | plus(activate(X1), activate(X2)) |
| activate(n__s(X)) | → | s(activate(X)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| isNat#(n__plus(V1, V2)) | → | isNat#(activate(V1)) | | plus#(N, 0) | → | isNat#(N) |
| isNat#(n__plus(V1, V2)) | → | and#(isNat(activate(V1)), n__isNat(activate(V2))) | | activate#(n__plus(X1, X2)) | → | activate#(X2) |
| plus#(N, 0) | → | U11#(isNat(N), N) | | activate#(n__isNat(X)) | → | isNat#(X) |
| activate#(n__plus(X1, X2)) | → | activate#(X1) |
Problem 4: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| and#(tt, X) | → | activate#(X) | | U11#(tt, N) | → | activate#(N) |
| activate#(n__plus(X1, X2)) | → | plus#(activate(X1), activate(X2)) | | isNat#(n__plus(V1, V2)) | → | activate#(V1) |
| isNat#(n__plus(V1, V2)) | → | activate#(V2) |
Rewrite Rules
| U11(tt, N) | → | activate(N) | | U21(tt, M, N) | → | s(plus(activate(N), activate(M))) |
| and(tt, X) | → | activate(X) | | isNat(n__0) | → | tt |
| isNat(n__plus(V1, V2)) | → | and(isNat(activate(V1)), n__isNat(activate(V2))) | | isNat(n__s(V1)) | → | isNat(activate(V1)) |
| plus(N, 0) | → | U11(isNat(N), N) | | plus(N, s(M)) | → | U21(and(isNat(M), n__isNat(N)), M, N) |
| 0 | → | n__0 | | plus(X1, X2) | → | n__plus(X1, X2) |
| isNat(X) | → | n__isNat(X) | | s(X) | → | n__s(X) |
| activate(n__0) | → | 0 | | activate(n__plus(X1, X2)) | → | plus(activate(X1), activate(X2)) |
| activate(n__isNat(X)) | → | isNat(X) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: plus, n__isNat, n__plus, and, activate, isNat, n__s, n__0, 0, s, tt, U11, U21
Strategy
There are no SCCs!