Certification Problem

Input (TPDB TRS_Standard/Mixed_TRS/test1)

The rewrite relation of the following TRS is considered.

f(s(x),y) f(x,s(s(x))) (1)
f(x,s(s(y))) f(y,x) (2)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
f#(s(x),y) f#(x,s(s(x))) (3)
f#(x,s(s(y))) f#(y,x) (4)

1.1 Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the arctic semiring over the naturals
[f#(x1, x2)] =
0
+
3
· x1 +
1
· x2
[s(x1)] =
2
+
1
· x1
having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
f#(s(x),y) f#(x,s(s(x))) (3)
could be deleted.

1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals
[s(x1)] = 1 · x1
[f#(x1, x2)] = 1 · x1 + 1 · x2
having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

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.

f#(x,s(s(y))) f#(y,x) (4)
2 > 1
1 2

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