Term Rewriting System R:
[X, X1, X2]
af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
Innermost Termination of R to be shown.
R
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
af(0) -> cons(0, f(s(0)))
af(X) -> f(X)
where the Polynomial interpretation:
| POL(a__p(x1)) | = x1 |
| POL(0) | = 0 |
| POL(cons(x1, x2)) | = x1 + x2 |
| POL(s(x1)) | = x1 |
| POL(mark(x1)) | = 2·x1 |
| POL(f(x1)) | = 1 + x1 |
| POL(p(x1)) | = x1 |
| POL(a__f(x1)) | = 2 + x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
mark(s(X)) -> s(mark(X))
ap(s(0)) -> 0
where the Polynomial interpretation:
| POL(a__p(x1)) | = x1 |
| POL(0) | = 0 |
| POL(cons(x1, x2)) | = x1 + x2 |
| POL(s(x1)) | = 1 + x1 |
| POL(mark(x1)) | = 2·x1 |
| POL(f(x1)) | = x1 |
| POL(p(x1)) | = x1 |
| POL(a__f(x1)) | = x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
mark(f(X)) -> af(mark(X))
where the Polynomial interpretation:
| POL(a__p(x1)) | = x1 |
| POL(0) | = 0 |
| POL(cons(x1, x2)) | = x1 + x2 |
| POL(s(x1)) | = x1 |
| POL(mark(x1)) | = x1 |
| POL(f(x1)) | = 1 + x1 |
| POL(p(x1)) | = x1 |
| POL(a__f(x1)) | = x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳RRRPolo
...
→TRS4
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
mark(0) -> 0
where the Polynomial interpretation:
| POL(a__p(x1)) | = x1 |
| POL(0) | = 1 |
| POL(cons(x1, x2)) | = x1 + x2 |
| POL(s(x1)) | = x1 |
| POL(mark(x1)) | = 2·x1 |
| POL(p(x1)) | = x1 |
| POL(a__f(x1)) | = x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳RRRPolo
...
→TRS5
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
mark(cons(X1, X2)) -> cons(mark(X1), X2)
where the Polynomial interpretation:
| POL(a__p(x1)) | = x1 |
| POL(0) | = 0 |
| POL(cons(x1, x2)) | = 1 + x1 + x2 |
| POL(s(x1)) | = x1 |
| POL(mark(x1)) | = 2·x1 |
| POL(p(x1)) | = x1 |
| POL(a__f(x1)) | = x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳RRRPolo
...
→TRS6
↳Dependency Pair Analysis
R contains the following Dependency Pairs:
MARK(p(X)) -> AP(mark(X))
MARK(p(X)) -> MARK(X)
AF(s(0)) -> AF(ap(s(0)))
AF(s(0)) -> AP(s(0))
Furthermore, R contains one SCC.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳RRRPolo
...
→DP Problem 1
↳Usable Rules (Innermost)
Dependency Pair:
MARK(p(X)) -> MARK(X)
Rules:
mark(p(X)) -> ap(mark(X))
ap(X) -> p(X)
af(s(0)) -> af(ap(s(0)))
Strategy:
innermost
As we are in the innermost case, we can delete all 3 non-usable-rules.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳RRRPolo
...
→DP Problem 2
↳Size-Change Principle
Dependency Pair:
MARK(p(X)) -> MARK(X)
Rule:
none
Strategy:
innermost
We number the DPs as follows:
- MARK(p(X)) -> MARK(X)
and get the following Size-Change Graph(s):
which lead(s) to this/these maximal multigraph(s):
DP: empty set
Oriented Rules: none
We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial
with Argument Filtering System:
p(x1) -> p(x1)
We obtain no new DP problems.
Innermost Termination of R successfully shown.
Duration:
0:00 minutes