Certification Problem

Input (TPDB SRS_Standard/Mixed_SRS/09)

The rewrite relation of the following TRS is considered.

a(b(a(x1))) → b(b(b(a(x1)))) (1)
b(b(b(a(x1)))) → a(a(a(b(x1)))) (2)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(b(a(x1))) → a(b(b(b(x1)))) (3)
a(b(b(b(x1)))) → b(a(a(a(x1)))) (4)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(b(a(x1))) → a#(b(b(b(x1)))) (5)
a#(b(b(b(x1)))) → a#(x1) (6)
a#(b(b(b(x1)))) → a#(a(x1)) (7)
a#(b(b(b(x1)))) → a#(a(a(x1))) (8)

1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[b(x1)] =
0 0 1
1 0 0
0 1 0
· x1 +
1 0 0
0 0 0
0 0 0
[a(x1)] =
0 0 1
0 0 1
0 0 1
· x1 +
0 0 0
1 0 0
0 0 0
[a#(x1)] =
0 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
together with the usable rules
a(b(a(x1))) → a(b(b(b(x1)))) (3)
a(b(b(b(x1)))) → b(a(a(a(x1)))) (4)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
a#(b(b(b(x1)))) → a#(x1) (6)
a#(b(b(b(x1)))) → a#(a(x1)) (7)
a#(b(b(b(x1)))) → a#(a(a(x1))) (8)
could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.