Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z048)

The rewrite relation of the following TRS is considered.

b(a(b(a(c(b(a(x1)))))))

→

a(b(a(c(b(b(a(b(a(c(x1))))))))))

(1)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS

a(b(c(a(b(a(b(x1)))))))

→

c(a(b(a(b(b(c(a(b(a(x1))))))))))

(2)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

a^#(b(c(a(b(a(b(x1)))))))	→	a^#(b(a(b(b(c(a(b(a(x1)))))))))	(3)
a^#(b(c(a(b(a(b(x1)))))))	→	a^#(b(b(c(a(b(a(x1)))))))	(4)
a^#(b(c(a(b(a(b(x1)))))))	→	a^#(b(a(x1)))	(5)
a^#(b(c(a(b(a(b(x1)))))))	→	a^#(x1)	(6)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(b(c(a(b(a(b(x1))))))) → a^#(x1) (6)

a^#(b(c(a(b(a(b(x1))))))) → a^#(b(a(x1))) (5)

1.1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals

[a^#(x₁)] = -2 + 2 · x₁

[b(x₁)] = 2 + x₁

[a(x₁)] = x₁

[c(x₁)] = -2 + x₁

the pairs

a^#(b(c(a(b(a(b(x1))))))) → a^#(x1) (6)

a^#(b(c(a(b(a(b(x1))))))) → a^#(b(a(x1))) (5)

could be deleted.
1.1.1.1.1 P is empty

There are no pairs anymore.