Certification Problem

Input (TPDB SRS_Standard/Mixed_SRS/04)

The rewrite relation of the following TRS is considered.

a(a(a(x1))) → a(b(b(b(x1)))) (1)
b(b(a(b(x1)))) → a(a(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(x1))) → b#(x1) (3)
a#(a(a(x1))) → b#(b(x1)) (4)
a#(a(a(x1))) → b#(b(b(x1))) (5)
a#(a(a(x1))) → a#(b(b(b(x1)))) (6)
b#(b(a(b(x1)))) → a#(x1) (7)
b#(b(a(b(x1)))) → a#(a(x1)) (8)
b#(b(a(b(x1)))) → a#(a(a(x1))) (9)

1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals
[b(x1)] = 1 · x1 + 2
[b#(x1)] = 4 · x1 + 3
[a(x1)] = 1 · x1 + 3
[a#(x1)] = 4 · x1 + 0
together with the usable rules
a(a(a(x1))) → a(b(b(b(x1)))) (1)
b(b(a(b(x1)))) → a(a(a(x1))) (2)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
a#(a(a(x1))) → b#(x1) (3)
a#(a(a(x1))) → b#(b(x1)) (4)
a#(a(a(x1))) → b#(b(b(x1))) (5)
b#(b(a(b(x1)))) → a#(x1) (7)
b#(b(a(b(x1)))) → a#(a(x1)) (8)
b#(b(a(b(x1)))) → a#(a(a(x1))) (9)
could be deleted.

1.1.1 Reduction Pair Processor with Usable Rules

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

1.1.1.1 P is empty

There are no pairs anymore.