Certification Problem

Input (TPDB SRS_Standard/Mixed_SRS/07)

The rewrite relation of the following TRS is considered.

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

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(b(a(a(x1)))) → b#(a(x1)) (4)
b#(b(a(a(x1)))) → a#(b(a(x1))) (5)
b#(b(a(a(x1)))) → a#(a(b(a(x1)))) (6)
a#(x1) → b#(x1) (7)
a#(x1) → b#(b(x1)) (8)
a#(x1) → b#(b(b(x1))) (9)

1.1.1 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)] =
1 0
0 -∞
· x1 +
1 -∞
0 -∞
[a#(x1)] =
1 0
-∞ -∞
· x1 +
1 -∞
-∞ -∞
[b(x1)] =
-∞ 0
0 -∞
· x1 +
0 -∞
0 -∞
[b#(x1)] =
-∞ 0
-∞ -∞
· x1 +
0 -∞
-∞ -∞
together with the usable rules
b(b(a(a(x1)))) → a(a(b(a(x1)))) (3)
a(x1) → b(b(b(x1))) (2)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
b#(b(a(a(x1)))) → b#(a(x1)) (4)
b#(b(a(a(x1)))) → a#(b(a(x1))) (5)
a#(x1) → b#(b(x1)) (8)
could be deleted.

1.1.1.1 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 1
· x1 +
1 -∞
2 -∞
[a#(x1)] =
1 1
-∞ -∞
· x1 +
1 -∞
-∞ -∞
[b(x1)] =
-∞ 0
0 -∞
· x1 +
0 -∞
0 -∞
[b#(x1)] =
0 0
-∞ -∞
· x1 +
0 -∞
-∞ -∞
together with the usable rules
b(b(a(a(x1)))) → a(a(b(a(x1)))) (3)
a(x1) → b(b(b(x1))) (2)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
a#(x1) → b#(x1) (7)
a#(x1) → b#(b(b(x1))) (9)
could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.