Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/torpa2)

The rewrite relation of the following TRS is considered.

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

1.1 Dependency Pair Transformation

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

1.1.1 Subterm Criterion Processor

We use the projection to multisets
π(a#) = { 1, 1 }
π(b#) = { 1, 1 }
π(a) = { 1, 1 }
π(b) = { 1, 1 }
to remove the pairs:
b#(a(b(x1))) → a#(x1) (4)
b#(a(b(x1))) → b#(a(x1)) (5)

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

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.