Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-97)

The rewrite relation of the following TRS is considered.

a(a(a(b(x1)))) b(a(b(a(x1)))) (1)
a(a(b(b(x1)))) a(a(a(b(x1)))) (2)
a(a(a(b(x1)))) b(b(a(b(x1)))) (3)
b(b(b(a(x1)))) b(a(b(b(x1)))) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers
[a(x1)] =
0 2
0 1
· x1 +
-∞ -∞
-∞ -∞
[b(x1)] =
1 1
-∞ -∞
· x1 +
-∞ -∞
-∞ -∞
all of the following rules can be deleted.
a(a(a(b(x1)))) b(b(a(b(x1)))) (3)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(a(a(b(x1)))) a#(x1) (5)
a#(a(a(b(x1)))) b#(a(x1)) (6)
a#(a(a(b(x1)))) a#(b(a(x1))) (7)
a#(a(a(b(x1)))) b#(a(b(a(x1)))) (8)
a#(a(b(b(x1)))) a#(b(x1)) (9)
a#(a(b(b(x1)))) a#(a(b(x1))) (10)
a#(a(b(b(x1)))) a#(a(a(b(x1)))) (11)
b#(b(b(a(x1)))) b#(x1) (12)
b#(b(b(a(x1)))) b#(b(x1)) (13)
b#(b(b(a(x1)))) a#(b(b(x1))) (14)
b#(b(b(a(x1)))) b#(a(b(b(x1)))) (15)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.