YES
The TRS could be proven terminating. The proof took 885 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (115ms).
| Problem 2 was processed with processor PolynomialLinearRange4iUR (222ms).
| | Problem 3 was processed with processor DependencyGraph (13ms).
| | | Problem 4 was processed with processor PolynomialLinearRange4iUR (81ms).
| | | | Problem 5 was processed with processor DependencyGraph (3ms).
| | | | | Problem 6 was processed with processor PolynomialLinearRange4iUR (68ms).
| | | | | | Problem 7 was processed with processor DependencyGraph (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
add#(s(X), Y) | → | activate#(X) | | add#(s(X), Y) | → | activate#(Y) |
activate#(n__from(X)) | → | from#(X) | | activate#(n__s(X)) | → | s#(X) |
add#(s(X), Y) | → | s#(n__add(activate(X), activate(Y))) | | first#(s(X), cons(Y, Z)) | → | activate#(X) |
activate#(n__add(X1, X2)) | → | add#(X1, X2) | | from#(X) | → | activate#(X) |
first#(s(X), cons(Y, Z)) | → | activate#(Z) | | if#(true, X, Y) | → | activate#(X) |
if#(false, X, Y) | → | activate#(Y) | | and#(true, X) | → | activate#(X) |
first#(s(X), cons(Y, Z)) | → | activate#(Y) | | add#(0, X) | → | activate#(X) |
activate#(n__first(X1, X2)) | → | first#(X1, X2) |
Rewrite Rules
and(true, X) | → | activate(X) | | and(false, Y) | → | false |
if(true, X, Y) | → | activate(X) | | if(false, X, Y) | → | activate(Y) |
add(0, X) | → | activate(X) | | add(s(X), Y) | → | s(n__add(activate(X), activate(Y))) |
first(0, X) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(activate(Y), n__first(activate(X), activate(Z))) |
from(X) | → | cons(activate(X), n__from(n__s(activate(X)))) | | add(X1, X2) | → | n__add(X1, X2) |
first(X1, X2) | → | n__first(X1, X2) | | from(X) | → | n__from(X) |
s(X) | → | n__s(X) | | activate(n__add(X1, X2)) | → | add(X1, X2) |
activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(n__from(X)) | → | from(X) |
activate(n__s(X)) | → | s(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, true, from, add, and, n__s, activate, 0, s, n__first, if, n__add, false, first, cons, nil
Strategy
The following SCCs where found
activate#(n__add(X1, X2)) → add#(X1, X2) | from#(X) → activate#(X) |
first#(s(X), cons(Y, Z)) → activate#(Z) | add#(s(X), Y) → activate#(X) |
add#(s(X), Y) → activate#(Y) | first#(s(X), cons(Y, Z)) → activate#(Y) |
add#(0, X) → activate#(X) | activate#(n__from(X)) → from#(X) |
activate#(n__first(X1, X2)) → first#(X1, X2) | first#(s(X), cons(Y, Z)) → activate#(X) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
activate#(n__add(X1, X2)) | → | add#(X1, X2) | | from#(X) | → | activate#(X) |
first#(s(X), cons(Y, Z)) | → | activate#(Z) | | add#(s(X), Y) | → | activate#(X) |
add#(s(X), Y) | → | activate#(Y) | | first#(s(X), cons(Y, Z)) | → | activate#(Y) |
add#(0, X) | → | activate#(X) | | activate#(n__from(X)) | → | from#(X) |
activate#(n__first(X1, X2)) | → | first#(X1, X2) | | first#(s(X), cons(Y, Z)) | → | activate#(X) |
Rewrite Rules
and(true, X) | → | activate(X) | | and(false, Y) | → | false |
if(true, X, Y) | → | activate(X) | | if(false, X, Y) | → | activate(Y) |
add(0, X) | → | activate(X) | | add(s(X), Y) | → | s(n__add(activate(X), activate(Y))) |
first(0, X) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(activate(Y), n__first(activate(X), activate(Z))) |
from(X) | → | cons(activate(X), n__from(n__s(activate(X)))) | | add(X1, X2) | → | n__add(X1, X2) |
first(X1, X2) | → | n__first(X1, X2) | | from(X) | → | n__from(X) |
s(X) | → | n__s(X) | | activate(n__add(X1, X2)) | → | add(X1, X2) |
activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(n__from(X)) | → | from(X) |
activate(n__s(X)) | → | s(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, true, from, add, and, n__s, activate, 0, s, n__first, if, n__add, false, first, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): 0
- activate#(x): x + 1
- add(x,y): 0
- add#(x,y): y + x + 1
- and(x,y): 0
- cons(x,y): y + x
- false: 0
- first(x,y): 0
- first#(x,y): 2y + 2x + 2
- from(x): 0
- from#(x): x + 1
- if(x,y,z): 0
- n__add(x,y): 2y + x
- n__first(x,y): 2y + 3x + 1
- n__from(x): 2x
- n__s(x): 0
- nil: 0
- s(x): x
- true: 0
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
first#(s(X), cons(Y, Z)) | → | activate#(Z) | | first#(s(X), cons(Y, Z)) | → | activate#(Y) |
first#(s(X), cons(Y, Z)) | → | activate#(X) |
Problem 3: DependencyGraph
Dependency Pair Problem
Dependency Pairs
activate#(n__add(X1, X2)) | → | add#(X1, X2) | | from#(X) | → | activate#(X) |
add#(s(X), Y) | → | activate#(X) | | add#(s(X), Y) | → | activate#(Y) |
add#(0, X) | → | activate#(X) | | activate#(n__from(X)) | → | from#(X) |
activate#(n__first(X1, X2)) | → | first#(X1, X2) |
Rewrite Rules
and(true, X) | → | activate(X) | | and(false, Y) | → | false |
if(true, X, Y) | → | activate(X) | | if(false, X, Y) | → | activate(Y) |
add(0, X) | → | activate(X) | | add(s(X), Y) | → | s(n__add(activate(X), activate(Y))) |
first(0, X) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(activate(Y), n__first(activate(X), activate(Z))) |
from(X) | → | cons(activate(X), n__from(n__s(activate(X)))) | | add(X1, X2) | → | n__add(X1, X2) |
first(X1, X2) | → | n__first(X1, X2) | | from(X) | → | n__from(X) |
s(X) | → | n__s(X) | | activate(n__add(X1, X2)) | → | add(X1, X2) |
activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(n__from(X)) | → | from(X) |
activate(n__s(X)) | → | s(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, true, from, add, and, n__s, activate, 0, s, n__first, if, n__add, false, first, nil, cons
Strategy
The following SCCs where found
activate#(n__add(X1, X2)) → add#(X1, X2) | from#(X) → activate#(X) |
add#(s(X), Y) → activate#(X) | add#(s(X), Y) → activate#(Y) |
add#(0, X) → activate#(X) | activate#(n__from(X)) → from#(X) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
activate#(n__add(X1, X2)) | → | add#(X1, X2) | | from#(X) | → | activate#(X) |
add#(s(X), Y) | → | activate#(X) | | add#(s(X), Y) | → | activate#(Y) |
add#(0, X) | → | activate#(X) | | activate#(n__from(X)) | → | from#(X) |
Rewrite Rules
and(true, X) | → | activate(X) | | and(false, Y) | → | false |
if(true, X, Y) | → | activate(X) | | if(false, X, Y) | → | activate(Y) |
add(0, X) | → | activate(X) | | add(s(X), Y) | → | s(n__add(activate(X), activate(Y))) |
first(0, X) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(activate(Y), n__first(activate(X), activate(Z))) |
from(X) | → | cons(activate(X), n__from(n__s(activate(X)))) | | add(X1, X2) | → | n__add(X1, X2) |
first(X1, X2) | → | n__first(X1, X2) | | from(X) | → | n__from(X) |
s(X) | → | n__s(X) | | activate(n__add(X1, X2)) | → | add(X1, X2) |
activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(n__from(X)) | → | from(X) |
activate(n__s(X)) | → | s(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, true, from, add, and, n__s, activate, 0, s, n__first, if, n__add, false, first, nil, cons
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): 0
- activate#(x): x
- add(x,y): 0
- add#(x,y): y + x
- and(x,y): 0
- cons(x,y): 0
- false: 0
- first(x,y): 0
- from(x): 0
- from#(x): x + 1
- if(x,y,z): 0
- n__add(x,y): 2y + 2x
- n__first(x,y): 0
- n__from(x): x + 1
- n__s(x): 0
- nil: 0
- s(x): x
- true: 0
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
Problem 5: DependencyGraph
Dependency Pair Problem
Dependency Pairs
activate#(n__add(X1, X2)) | → | add#(X1, X2) | | add#(s(X), Y) | → | activate#(X) |
add#(s(X), Y) | → | activate#(Y) | | add#(0, X) | → | activate#(X) |
activate#(n__from(X)) | → | from#(X) |
Rewrite Rules
and(true, X) | → | activate(X) | | and(false, Y) | → | false |
if(true, X, Y) | → | activate(X) | | if(false, X, Y) | → | activate(Y) |
add(0, X) | → | activate(X) | | add(s(X), Y) | → | s(n__add(activate(X), activate(Y))) |
first(0, X) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(activate(Y), n__first(activate(X), activate(Z))) |
from(X) | → | cons(activate(X), n__from(n__s(activate(X)))) | | add(X1, X2) | → | n__add(X1, X2) |
first(X1, X2) | → | n__first(X1, X2) | | from(X) | → | n__from(X) |
s(X) | → | n__s(X) | | activate(n__add(X1, X2)) | → | add(X1, X2) |
activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(n__from(X)) | → | from(X) |
activate(n__s(X)) | → | s(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, true, from, add, and, n__s, activate, 0, s, n__first, if, n__add, false, first, cons, nil
Strategy
The following SCCs where found
activate#(n__add(X1, X2)) → add#(X1, X2) | add#(s(X), Y) → activate#(X) |
add#(s(X), Y) → activate#(Y) | add#(0, X) → activate#(X) |
Problem 6: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
activate#(n__add(X1, X2)) | → | add#(X1, X2) | | add#(s(X), Y) | → | activate#(X) |
add#(s(X), Y) | → | activate#(Y) | | add#(0, X) | → | activate#(X) |
Rewrite Rules
and(true, X) | → | activate(X) | | and(false, Y) | → | false |
if(true, X, Y) | → | activate(X) | | if(false, X, Y) | → | activate(Y) |
add(0, X) | → | activate(X) | | add(s(X), Y) | → | s(n__add(activate(X), activate(Y))) |
first(0, X) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(activate(Y), n__first(activate(X), activate(Z))) |
from(X) | → | cons(activate(X), n__from(n__s(activate(X)))) | | add(X1, X2) | → | n__add(X1, X2) |
first(X1, X2) | → | n__first(X1, X2) | | from(X) | → | n__from(X) |
s(X) | → | n__s(X) | | activate(n__add(X1, X2)) | → | add(X1, X2) |
activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(n__from(X)) | → | from(X) |
activate(n__s(X)) | → | s(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, true, from, add, and, n__s, activate, 0, s, n__first, if, n__add, false, first, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): 0
- activate#(x): x
- add(x,y): 0
- add#(x,y): 2y + 2x
- and(x,y): 0
- cons(x,y): 0
- false: 0
- first(x,y): 0
- from(x): 0
- if(x,y,z): 0
- n__add(x,y): 2y + 2x + 2
- n__first(x,y): 0
- n__from(x): 0
- n__s(x): 0
- nil: 0
- s(x): 2x
- true: 0
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
activate#(n__add(X1, X2)) | → | add#(X1, X2) |
Problem 7: DependencyGraph
Dependency Pair Problem
Dependency Pairs
add#(s(X), Y) | → | activate#(X) | | add#(s(X), Y) | → | activate#(Y) |
add#(0, X) | → | activate#(X) |
Rewrite Rules
and(true, X) | → | activate(X) | | and(false, Y) | → | false |
if(true, X, Y) | → | activate(X) | | if(false, X, Y) | → | activate(Y) |
add(0, X) | → | activate(X) | | add(s(X), Y) | → | s(n__add(activate(X), activate(Y))) |
first(0, X) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(activate(Y), n__first(activate(X), activate(Z))) |
from(X) | → | cons(activate(X), n__from(n__s(activate(X)))) | | add(X1, X2) | → | n__add(X1, X2) |
first(X1, X2) | → | n__first(X1, X2) | | from(X) | → | n__from(X) |
s(X) | → | n__s(X) | | activate(n__add(X1, X2)) | → | add(X1, X2) |
activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(n__from(X)) | → | from(X) |
activate(n__s(X)) | → | s(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, true, from, add, and, n__s, activate, 0, s, n__first, if, n__add, false, first, nil, cons
Strategy
There are no SCCs!