Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z080)

The rewrite relation of the following TRS is considered.

A(b(x1)) → b(a(B(A(x1)))) (1)
B(a(x1)) → a(b(A(B(x1)))) (2)
A(a(x1)) → x1 (3)
B(b(x1)) → x1 (4)

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#(b(x1)) → A#(x1) (5)
A#(b(x1)) → B#(A(x1)) (6)
B#(a(x1)) → B#(x1) (7)
B#(a(x1)) → A#(B(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 0
-∞ -∞
· x1 +
0 -∞
-∞ -∞
[A(x1)] =
-∞ 0
-∞ 0
· x1 +
0 -∞
0 -∞
[a(x1)] =
1 0
0 0
· x1 +
1 -∞
0 -∞
[B#(x1)] =
-∞ 0
-∞ -∞
· x1 +
0 -∞
-∞ -∞
[b(x1)] =
-∞ 0
0 1
· x1 +
0 -∞
1 -∞
[B(x1)] =
0 0
0 -∞
· x1 +
-∞ -∞
0 -∞
together with the usable rules
A(b(x1)) → b(a(B(A(x1)))) (1)
B(a(x1)) → a(b(A(B(x1)))) (2)
A(a(x1)) → x1 (3)
B(b(x1)) → x1 (4)
(w.r.t. the implicit argument filter of the reduction pair), the pair
A#(b(x1)) → B#(A(x1)) (6)
could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.