Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z030)

The rewrite relation of the following TRS is considered.

b(a(b(a(a(a(x1)))))) → a(a(a(b(a(b(a(b(a(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(b(a(a(a(x1))))))) → b(a(a(a(b(a(b(a(b(a(x1)))))))))) (2)
a(b(a(b(a(a(a(x1))))))) → a(a(a(a(b(a(b(a(b(a(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(a0(b1(a1(a1(a0(x1))))))) → b1(a1(a1(a0(b1(a0(b1(a0(b1(a0(x1)))))))))) (4)
b0(b1(a0(b1(a1(a1(a1(x1))))))) → b1(a1(a1(a0(b1(a0(b1(a0(b1(a1(x1)))))))))) (5)
a0(b1(a0(b1(a1(a1(a0(x1))))))) → a1(a1(a1(a0(b1(a0(b1(a0(b1(a0(x1)))))))))) (6)
a0(b1(a0(b1(a1(a1(a1(x1))))))) → a1(a1(a1(a0(b1(a0(b1(a0(b1(a1(x1)))))))))) (7)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[b0(x1)] = x1 +
1
[b1(x1)] = x1 +
0
[a0(x1)] = x1 +
0
[a1(x1)] = x1 +
0
all of the following rules can be deleted.
b0(b1(a0(b1(a1(a1(a0(x1))))))) → b1(a1(a1(a0(b1(a0(b1(a0(b1(a0(x1)))))))))) (4)
b0(b1(a0(b1(a1(a1(a1(x1))))))) → b1(a1(a1(a0(b1(a0(b1(a0(b1(a1(x1)))))))))) (5)

1.1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a0(a1(a1(b1(a0(b1(a0(x1))))))) → a0(b1(a0(b1(a0(b1(a0(a1(a1(a1(x1)))))))))) (8)
a1(a1(a1(b1(a0(b1(a0(x1))))))) → a1(b1(a0(b1(a0(b1(a0(a1(a1(a1(x1)))))))))) (9)

1.1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a0#(a1(a1(b1(a0(b1(a0(x1))))))) → a0#(b1(a0(b1(a0(b1(a0(a1(a1(a1(x1)))))))))) (10)
a0#(a1(a1(b1(a0(b1(a0(x1))))))) → a0#(b1(a0(b1(a0(a1(a1(a1(x1)))))))) (11)
a0#(a1(a1(b1(a0(b1(a0(x1))))))) → a0#(b1(a0(a1(a1(a1(x1)))))) (12)
a0#(a1(a1(b1(a0(b1(a0(x1))))))) → a0#(a1(a1(a1(x1)))) (13)
a0#(a1(a1(b1(a0(b1(a0(x1))))))) → a1#(x1) (14)
a0#(a1(a1(b1(a0(b1(a0(x1))))))) → a1#(a1(x1)) (15)
a0#(a1(a1(b1(a0(b1(a0(x1))))))) → a1#(a1(a1(x1))) (16)
a1#(a1(a1(b1(a0(b1(a0(x1))))))) → a0#(b1(a0(b1(a0(a1(a1(a1(x1)))))))) (17)
a1#(a1(a1(b1(a0(b1(a0(x1))))))) → a0#(b1(a0(a1(a1(a1(x1)))))) (18)
a1#(a1(a1(b1(a0(b1(a0(x1))))))) → a0#(a1(a1(a1(x1)))) (19)
a1#(a1(a1(b1(a0(b1(a0(x1))))))) → a1#(x1) (20)
a1#(a1(a1(b1(a0(b1(a0(x1))))))) → a1#(b1(a0(b1(a0(b1(a0(a1(a1(a1(x1)))))))))) (21)
a1#(a1(a1(b1(a0(b1(a0(x1))))))) → a1#(a1(x1)) (22)
a1#(a1(a1(b1(a0(b1(a0(x1))))))) → a1#(a1(a1(x1))) (23)

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.