Certification Problem

Input (TPDB TRS_Standard/Waldmann_06/jwmatchb2)

The rewrite relation of the following TRS is considered.

h(f(x,y)) → f(f(a,h(h(y))),x) (1)

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:

10

Hence, it suffices to show innermost termination in the following.

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
h#(f(x,y)) → h#(h(y)) (2)
h#(f(x,y)) → h#(y) (3)

1.1.1 Narrowing Processor

We consider all narrowings of the pair below position 1 to get the following set of pairs
h#(f(y0,f(x0,x1))) → h#(f(f(a,h(h(x1))),x0)) (4)

1.1.1.1 Forward Instantiation Processor

We instantiate the pair to the following set of pairs
h#(f(x0,f(y_0,y_1))) → h#(f(y_0,y_1)) (5)
h#(f(x0,f(y_0,f(y_1,y_2)))) → h#(f(y_0,f(y_1,y_2))) (6)

1.1.1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
[h#(x1)] =
0
-∞
-∞
+
-∞ -∞ 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[f(x1, x2)] =
0
1
1
+
-∞ -∞ -∞
1 1 0
0 0 -∞
· x1 +
-∞ -∞ -∞
-∞ -∞ -∞
1 1 0
· x2
[a] =
0
-∞
0
[h(x1)] =
1
0
-∞
+
-∞ -∞ -∞
0 -∞ 1
0 0 -∞
· x1
the pair
h#(f(y0,f(x0,x1))) → h#(f(f(a,h(h(x1))),x0)) (4)
could be deleted.

1.1.1.1.1.1 Usable Rules Processor

We restrict the rewrite rules to the following usable rules of the DP problem.

There are no rules.

1.1.1.1.1.1.1 Innermost Lhss Removal Processor

We restrict the innermost strategy to the following left hand sides.

There are no lhss.

1.1.1.1.1.1.1.1 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

h#(f(x0,f(y_0,y_1))) → h#(f(y_0,y_1)) (5)
1 > 1
h#(f(x0,f(y_0,f(y_1,y_2)))) → h#(f(y_0,f(y_1,y_2))) (6)
1 > 1

As there is no critical graph in the transitive closure, there are no infinite chains.