Term Rewriting System R:
[x, y]
ap(ap(g, x), y) -> y
ap(f, x) -> ap(f, app(g, x))
Innermost Termination of R to be shown.
R
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
ap(ap(g, x), y) -> y
where the Polynomial interpretation:
POL(g) | = 0 |
POL(ap(x1, x2)) | = 1 + x1 + x2 |
POL(f) | = 0 |
POL(app(x1, x2)) | = x1 + x2 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳Dependency Pair Analysis
R contains the following Dependency Pairs:
AP(f, x) -> AP(f, app(g, x))
Furthermore, R contains one SCC.
R
↳RRRPolo
→TRS2
↳DPs
→DP Problem 1
↳Usable Rules (Innermost)
Dependency Pair:
AP(f, x) -> AP(f, app(g, x))
Rule:
ap(f, x) -> ap(f, app(g, x))
Strategy:
innermost
As we are in the innermost case, we can delete all 1 non-usable-rules.
R
↳RRRPolo
→TRS2
↳DPs
→DP Problem 1
↳UsableRules
...
→DP Problem 2
↳A-Transformation
Dependency Pair:
AP(f, x) -> AP(f, app(g, x))
Rule:
none
Strategy:
innermost
We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.
R
↳RRRPolo
→TRS2
↳DPs
→DP Problem 1
↳UsableRules
...
→DP Problem 3
↳Non Termination
Dependency Pair:
F(x) -> F(g(x))
Rule:
none
Strategy:
innermost
Found an infinite P-chain over R:
P =
F(x) -> F(g(x))
R = none
s = F(x)
evaluates to t =F(g(x))
Thus, s starts an infinite chain as s matches t.
Innermost Non-Termination of R could be shown.
Duration:
0:01 minutes