YES
Termination Proof using AProVETerm Rewriting System R:
[X, Y, Z, I, P]
((X, Y), Z) -> (X, (Y, Z))
(X, nil) -> X
(nil, X) -> X
U11(tt) -> U12(tt)
U12(tt) -> tt
isNePal((I, (P, I))) -> U11(tt)
Termination of R to be shown.
TRS
↳CSR to TRS Transformation
Lucas-Transformation successful.
Replacement map:
U12(1)
(1, 2)
isNePal(1)
U11(1)
Old CSR:
((X, Y), Z) -> (X, (Y, Z))
(X, nil) -> X
(nil, X) -> X
U11(tt) -> U12(tt)
U12(tt) -> tt
isNePal((I, (P, I))) -> U11(tt)
new TRS:
((X, Y), Z) -> (X, (Y, Z))
(X, nil) -> X
(nil, X) -> X
U11(tt) -> U12(tt)
U12(tt) -> tt
isNePal((I, (P, I))) -> U11(tt)
TRS
↳CSRtoTRS
→TRS1
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
((X, Y), Z) -> (X, (Y, Z))
(X, nil) -> X
(nil, X) -> X
isNePal((I, (P, I))) -> U11(tt)
where the Polynomial interpretation:
| POL(U12(x1)) | = x1 |
| POL(__(x1, x2)) | = 1 + 2·x1 + x2 |
| POL(tt) | = 0 |
| POL(nil) | = 1 |
| POL(isNePal(x1)) | = x1 |
| POL(U11(x1)) | = x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
TRS
↳CSRtoTRS
→TRS1
↳RRRPolo
→TRS2
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
U12(tt) -> tt
where the Polynomial interpretation:
| POL(U12(x1)) | = 1 + x1 |
| POL(tt) | = 0 |
| POL(U11(x1)) | = 1 + x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
TRS
↳CSRtoTRS
→TRS1
↳RRRPolo
→TRS2
↳RRRPolo
...
→TRS3
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
U11(tt) -> U12(tt)
where the Polynomial interpretation:
| POL(U12(x1)) | = x1 |
| POL(tt) | = 0 |
| POL(U11(x1)) | = 1 + x1 |
was used.
All Rules of R can be deleted.
TRS
↳CSRtoTRS
→TRS1
↳RRRPolo
→TRS2
↳RRRPolo
...
→TRS4
↳Overlay and local confluence Check
The TRS is overlay and locally confluent (all critical pairs are trivially joinable).Hence, we can switch to innermost.
TRS
↳CSRtoTRS
→TRS1
↳RRRPolo
→TRS2
↳RRRPolo
...
→TRS5
↳Dependency Pair Analysis
R contains no Dependency Pairs and therefore no SCCs.
Termination of R successfully shown.
Duration:
0:01 minutes