Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-404)

The rewrite relation of the following TRS is considered.

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

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

b(b(a(b(x1))))	→	a(a(b(b(x1))))	(5)
b(b(a(a(x1))))	→	a(a(b(a(x1))))	(6)
b(b(b(b(x1))))	→	a(a(a(b(x1))))	(7)
a(a(b(a(x1))))	→	a(b(b(a(x1))))	(8)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

b^#(b(a(b(x1))))	→	a^#(a(b(b(x1))))	(9)
b^#(b(a(b(x1))))	→	a^#(b(b(x1)))	(10)
b^#(b(a(b(x1))))	→	b^#(b(x1))	(11)
b^#(b(a(a(x1))))	→	a^#(a(b(a(x1))))	(12)
b^#(b(a(a(x1))))	→	a^#(b(a(x1)))	(13)
b^#(b(a(a(x1))))	→	b^#(a(x1))	(14)
b^#(b(b(b(x1))))	→	a^#(a(a(b(x1))))	(15)
b^#(b(b(b(x1))))	→	a^#(a(b(x1)))	(16)
b^#(b(b(b(x1))))	→	a^#(b(x1))	(17)
a^#(a(b(a(x1))))	→	a^#(b(b(a(x1))))	(18)
a^#(a(b(a(x1))))	→	b^#(b(a(x1)))	(19)

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(x1))))	→	a^#(b(b(a(x1))))	(18)
a^#(a(b(a(x1))))	→	b^#(b(a(x1)))	(19)
b^#(b(a(b(x1))))	→	a^#(a(b(b(x1))))	(9)
b^#(b(a(b(x1))))	→	a^#(b(b(x1)))	(10)
b^#(b(a(b(x1))))	→	b^#(b(x1))	(11)
b^#(b(a(a(x1))))	→	a^#(a(b(a(x1))))	(12)
b^#(b(a(a(x1))))	→	a^#(b(a(x1)))	(13)
b^#(b(b(b(x1))))	→	a^#(a(a(b(x1))))	(15)
b^#(b(b(b(x1))))	→	a^#(a(b(x1)))	(16)
b^#(b(b(b(x1))))	→	a^#(b(x1))	(17)

1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

[a^#(x₁)]	=	1 · x₁
[a(x₁)]	=	1 + 1 · x₁
[b(x₁)]	=	1 + 1 · x₁
[b^#(x₁)]	=	1 · x₁

the pairs

a^#(a(b(a(x1))))	→	b^#(b(a(x1)))	(19)
b^#(b(a(b(x1))))	→	a^#(b(b(x1)))	(10)
b^#(b(a(b(x1))))	→	b^#(b(x1))	(11)
b^#(b(a(a(x1))))	→	a^#(b(a(x1)))	(13)
b^#(b(b(b(x1))))	→	a^#(a(b(x1)))	(16)
b^#(b(b(b(x1))))	→	a^#(b(x1))	(17)

could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

→

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

(18)

1.1.1.1.1.1 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	-∞	-∞
-∞	-∞	-∞

· x₁

[b(x₁)]

-∞

-∞	1	0
-∞	-∞	0
0	-∞	0

· x₁

the pair

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

→

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

(18)

could be deleted.

1.1.1.1.1.1.1 P is empty

There are no pairs anymore.