Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z089)

The rewrite relation of the following TRS is considered.

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

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(b(a(x1))))	→	b(a(c(a(x1))))	(4)
c(a(x1))	→	a(c(c(x1)))	(5)
c(c(c(x1)))	→	b(c(b(x1)))	(3)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

a^#(b(b(a(x1))))	→	a^#(c(a(x1)))	(6)
a^#(b(b(a(x1))))	→	c^#(a(x1))	(7)
c^#(a(x1))	→	a^#(c(c(x1)))	(8)
c^#(a(x1))	→	c^#(c(x1))	(9)
c^#(a(x1))	→	c^#(x1)	(10)
c^#(c(c(x1)))	→	c^#(b(x1))	(11)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(b(b(a(x1))))	→	c^#(a(x1))	(7)
c^#(a(x1))	→	a^#(c(c(x1)))	(8)
a^#(b(b(a(x1))))	→	a^#(c(a(x1)))	(6)
c^#(a(x1))	→	c^#(c(x1))	(9)
c^#(a(x1))	→	c^#(x1)	(10)

1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

[a^#(x₁)]	=	1 + 4 · x₁
[b(x₁)]	=	1 · x₁
[a(x₁)]	=	1 + 4 · x₁
[c^#(x₁)]	=	4 + 1 · x₁
[c(x₁)]	=	1 · x₁

the pairs

c^#(a(x1))	→	a^#(c(c(x1)))	(8)
c^#(a(x1))	→	c^#(c(x1))	(9)
c^#(a(x1))	→	c^#(x1)	(10)

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

→

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

(6)

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

[b(x₁)]

-∞

-∞	0	0
-∞	-∞	-∞
-∞	0	1

· x₁

[a(x₁)]

0	0	0
0	0	0
0	0	0

· x₁

[c(x₁)]

-∞

0	0	0
0	0	-∞
-∞	0	-∞

· x₁

the pair

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

→

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

(6)

could be deleted.

1.1.1.1.1.1.1 P is empty

There are no pairs anymore.