Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/2)

The rewrite relation of the following TRS is considered.

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

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(a(b(x1)))) → a#(x1) (3)
a#(a(a(b(x1)))) → b#(a(x1)) (4)
a#(a(a(b(x1)))) → a#(b(a(x1))) (5)
a#(a(a(b(x1)))) → b#(a(b(a(x1)))) (6)
b#(b(a(x1))) → a#(b(a(x1))) (7)
b#(b(a(x1))) → a#(a(b(a(x1)))) (8)

1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the arctic semiring over the integers
[a(x1)] =
-∞ 0 -∞
0 0 0
0 -∞ -∞
· x1 +
0 -∞ -∞
0 -∞ -∞
0 -∞ -∞
[b#(x1)] =
-∞ -∞ 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1 +
1 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
[b(x1)] =
-∞ -∞ 0
-∞ -∞ 0
1 -∞ 0
· x1 +
0 -∞ -∞
0 -∞ -∞
1 -∞ -∞
[a#(x1)] =
0 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1 +
0 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
together with the usable rules
a(a(a(b(x1)))) → b(a(b(a(x1)))) (1)
b(b(a(x1))) → a(a(b(a(x1)))) (2)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
a#(a(a(b(x1)))) → a#(x1) (3)
a#(a(a(b(x1)))) → a#(b(a(x1))) (5)
could be deleted.

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 arctic semiring over the integers
[a(x1)] =
-∞ 0 -∞
-∞ -∞ 0
0 0 0
· x1 +
0 -∞ -∞
0 -∞ -∞
0 -∞ -∞
[b#(x1)] =
0 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1 +
0 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
[b(x1)] =
0 1 -∞
0 -∞ -∞
0 -∞ -∞
· x1 +
1 -∞ -∞
0 -∞ -∞
0 -∞ -∞
[a#(x1)] =
0 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1 +
0 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
together with the usable rules
a(a(a(b(x1)))) → b(a(b(a(x1)))) (1)
b(b(a(x1))) → a(a(b(a(x1)))) (2)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
a#(a(a(b(x1)))) → b#(a(x1)) (4)
a#(a(a(b(x1)))) → b#(a(b(a(x1)))) (6)
could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.