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 AProVE @ termCOMP 2023)

1 String Reversal

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

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

→

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

(2)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

a^#(a(b(a(a(a(b(x1)))))))	→	a^#(a(a(b(b(a(a(a(x1))))))))	(3)
a^#(a(b(a(a(a(b(x1)))))))	→	a^#(a(b(b(a(a(a(x1)))))))	(4)
a^#(a(b(a(a(a(b(x1)))))))	→	a^#(b(b(a(a(a(x1))))))	(5)
a^#(a(b(a(a(a(b(x1)))))))	→	a^#(a(a(x1)))	(6)
a^#(a(b(a(a(a(b(x1)))))))	→	a^#(a(x1))	(7)
a^#(a(b(a(a(a(b(x1)))))))	→	a^#(x1)	(8)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(a(b(a(a(a(b(x1)))))))	→	a^#(a(x1))	(7)
a^#(a(b(a(a(a(b(x1)))))))	→	a^#(a(a(x1)))	(6)
a^#(a(b(a(a(a(b(x1)))))))	→	a^#(x1)	(8)

1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

[a^#(x₁)]	=	-2 + 2 · x₁
[a(x₁)]	=	2 + x₁
[b(x₁)]	=	-2 + x₁

the pairs

a^#(a(b(a(a(a(b(x1)))))))	→	a^#(a(x1))	(7)
a^#(a(b(a(a(a(b(x1)))))))	→	a^#(x1)	(8)

could be deleted.

1.1.1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals

[a^#(x₁)]

-∞

-∞	0	0
-∞	-∞	-∞
-∞	-∞	-∞

· x₁

[a(x₁)]

-∞	0	0
0	-∞	0
-∞	-∞	0

· x₁

[b(x₁)]

-∞

-∞	1	0
-∞	-∞	-∞
-∞	-∞	0

· x₁

the pair

a^#(a(b(a(a(a(b(x1)))))))

→

a^#(a(a(x1)))

(6)

could be deleted.

1.1.1.1.1.1 P is empty

There are no pairs anymore.