YES
Termination Proof using AProVETerm Rewriting System R:
[X, N, M]
and(tt, X) -> X
plus(N, 0) -> N
plus(N, s(M)) -> s(plus(N, M))
Termination of R to be shown.
R
↳Orthogonal Check
This CSR is orthogonal, so it is sufficient to show innermost termination.
R
↳OrthoCSR
→TRS1
↳CSR to TRS Transformation
Zantema-Transformation successful.
Replacement map:
and(1, 2)
plus(1, 2)
s(1)
Old CSR:
and(tt, X) -> X
plus(N, 0) -> N
plus(N, s(M)) -> s(plus(N, M))
new TRS:
and(tt, X) -> a(X)
plus(N, 0) -> N
plus(N, s(M)) -> s(plus(N, M))
R
↳OrthoCSR
→TRS1
↳CSRtoTRS
→TRS2
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
and(tt, X) -> a(X)
where the Polynomial interpretation:
| POL(and(x1, x2)) | = x1 + x2 |
| POL(plus(x1, x2)) | = x1 + x2 |
| POL(0) | = 0 |
| POL(*a*(x1)) | = x1 |
| POL(tt) | = 1 |
| POL(s(x1)) | = x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳OrthoCSR
→TRS1
↳CSRtoTRS
→TRS2
↳RRRPolo
...
→TRS3
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
plus(N, 0) -> N
where the Polynomial interpretation:
| POL(plus(x1, x2)) | = x1 + x2 |
| POL(0) | = 1 |
| POL(s(x1)) | = x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳OrthoCSR
→TRS1
↳CSRtoTRS
→TRS2
↳RRRPolo
...
→TRS4
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
plus(N, s(M)) -> s(plus(N, M))
where the Polynomial interpretation:
| POL(plus(x1, x2)) | = x1 + 2·x2 |
| POL(s(x1)) | = 1 + x1 |
was used.
All Rules of R can be deleted.
R
↳OrthoCSR
→TRS1
↳CSRtoTRS
→TRS2
↳RRRPolo
...
→TRS5
↳Overlay and local confluence Check
The TRS is overlay and locally confluent (all critical pairs are trivially joinable).Hence, we can switch to innermost.
R
↳OrthoCSR
→TRS1
↳CSRtoTRS
→TRS2
↳RRRPolo
...
→TRS6
↳Dependency Pair Analysis
R contains no Dependency Pairs and therefore no SCCs.
Termination of R successfully shown.
Duration:
0:01 minutes