Certification Problem

Input (TPDB SRS_Standard/Bouchare_06/13)

The rewrite relation of the following TRS is considered.

a(a(b(x1))) a(x1) (1)
a(b(a(x1))) b(b(a(x1))) (2)
b(b(x1)) a(a(a(x1))) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(a(b(x1))) a#(x1) (4)
a#(b(a(x1))) b#(b(a(x1))) (5)
b#(b(x1)) a#(x1) (6)
b#(b(x1)) a#(a(x1)) (7)
b#(b(x1)) a#(a(a(x1))) (8)

1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers
[a(x1)] =
0 -∞
1 0
· x1 +
-∞ -∞
0 -∞
[b#(x1)] =
-∞ 2
-∞ -∞
· x1 +
0 -∞
-∞ -∞
[b(x1)] =
0 0
1 0
· x1 +
1 -∞
1 -∞
[a#(x1)] =
2 1
-∞ -∞
· x1 +
0 -∞
-∞ -∞
together with the usable rules
a(a(b(x1))) a(x1) (1)
a(b(a(x1))) b(b(a(x1))) (2)
b(b(x1)) a(a(a(x1))) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
b#(b(x1)) a#(x1) (6)
b#(b(x1)) a#(a(x1)) (7)
b#(b(x1)) a#(a(a(x1))) (8)
could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.