Certification Problem
Input (TPDB TRS_Standard/SK90/4.41)
The rewrite relation of the following TRS is considered.
|
f(a,y) |
→ |
f(y,g(y)) |
(1) |
|
g(a) |
→ |
b |
(2) |
|
g(b) |
→ |
b |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Switch to Innermost Termination
The TRS is overlay and locally confluent:
10Hence, it suffices to show innermost termination in the following.
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
f#(a,y) |
→ |
f#(y,g(y)) |
(4) |
|
f#(a,y) |
→ |
g#(y) |
(5) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.
-
The
1st
component contains the
pair
1.1.1.1 Usable Rules Processor
We restrict the rewrite rules to the following usable rules of the DP problem.
|
g(a) |
→ |
b |
(2) |
|
g(b) |
→ |
b |
(3) |
1.1.1.1.1 Innermost Lhss Removal Processor
We restrict the innermost strategy to the following left hand sides.
1.1.1.1.1.1 Reduction Pair Processor
Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(f#) |
= |
2 |
|
weight(f#) |
= |
1 |
|
|
|
| prec(a) |
= |
0 |
|
weight(a) |
= |
3 |
|
|
|
| prec(g) |
= |
3 |
|
weight(g) |
= |
2 |
|
|
|
| prec(b) |
= |
1 |
|
weight(b) |
= |
1 |
|
|
|
in combination with the following argument filter
| π(f#) |
= |
[1,2] |
| π(a) |
= |
[] |
| π(g) |
= |
[] |
| π(b) |
= |
[] |
the
pair
could be deleted.
1.1.1.1.1.1.1 P is empty
There are no pairs anymore.