Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z036)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{b(☐), a(☐)}

We obtain the transformed TRS
b(b(a(a(a(b(a(a(x1)))))))) → b(a(a(a(b(b(a(a(a(b(x1)))))))))) (2)
a(b(a(a(a(b(a(a(x1)))))))) → a(a(a(a(b(b(a(a(a(b(x1)))))))))) (3)

1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,1}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 2):

[b(x1)] = 2x1 + 0
[a(x1)] = 2x1 + 1

We obtain the labeled TRS
b0(b1(a1(a1(a0(b1(a1(a0(x1)))))))) → b1(a1(a1(a0(b0(b1(a1(a1(a0(b0(x1)))))))))) (4)
b0(b1(a1(a1(a0(b1(a1(a1(x1)))))))) → b1(a1(a1(a0(b0(b1(a1(a1(a0(b1(x1)))))))))) (5)
a0(b1(a1(a1(a0(b1(a1(a0(x1)))))))) → a1(a1(a1(a0(b0(b1(a1(a1(a0(b0(x1)))))))))) (6)
a0(b1(a1(a1(a0(b1(a1(a1(x1)))))))) → a1(a1(a1(a0(b0(b1(a1(a1(a0(b1(x1)))))))))) (7)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b0#(b1(a1(a1(a0(b1(a1(a0(x1)))))))) → b0#(x1) (8)
b0#(b1(a1(a1(a0(b1(a1(a0(x1)))))))) → b0#(b1(a1(a1(a0(b0(x1)))))) (9)
b0#(b1(a1(a1(a0(b1(a1(a0(x1)))))))) → a0#(b0(x1)) (10)
b0#(b1(a1(a1(a0(b1(a1(a0(x1)))))))) → a0#(b0(b1(a1(a1(a0(b0(x1))))))) (11)
b0#(b1(a1(a1(a0(b1(a1(a1(x1)))))))) → b0#(b1(a1(a1(a0(b1(x1)))))) (12)
b0#(b1(a1(a1(a0(b1(a1(a1(x1)))))))) → a0#(b0(b1(a1(a1(a0(b1(x1))))))) (13)
b0#(b1(a1(a1(a0(b1(a1(a1(x1)))))))) → a0#(b1(x1)) (14)
a0#(b1(a1(a1(a0(b1(a1(a0(x1)))))))) → b0#(x1) (15)
a0#(b1(a1(a1(a0(b1(a1(a0(x1)))))))) → b0#(b1(a1(a1(a0(b0(x1)))))) (16)
a0#(b1(a1(a1(a0(b1(a1(a0(x1)))))))) → a0#(b0(x1)) (17)
a0#(b1(a1(a1(a0(b1(a1(a0(x1)))))))) → a0#(b0(b1(a1(a1(a0(b0(x1))))))) (18)
a0#(b1(a1(a1(a0(b1(a1(a1(x1)))))))) → b0#(b1(a1(a1(a0(b1(x1)))))) (19)
a0#(b1(a1(a1(a0(b1(a1(a1(x1)))))))) → a0#(b0(b1(a1(a1(a0(b1(x1))))))) (20)
a0#(b1(a1(a1(a0(b1(a1(a1(x1)))))))) → a0#(b1(x1)) (21)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.