Certification Problem

Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-287)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

a^#(b(x1))	→	c^#(b(c(b(x1))))	(5)
a^#(b(x1))	→	b^#(c(b(x1)))	(6)
a^#(b(x1))	→	c^#(b(x1))	(7)
b^#(x1)	→	a^#(a(x1))	(8)
b^#(x1)	→	a^#(x1)	(9)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

1.1.1 Reduction Pair Processor

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

[a^#(x₁)]

-∞

-∞	-∞	1
-∞	-∞	-∞
-∞	-∞	-∞

· x₁

[b(x₁)]

-1

-∞

0	-1	2
-1	0	2
-1	0	2

· x₁

[b^#(x₁)]

-∞

-∞	-∞	2
-∞	-∞	-∞
-∞	-∞	-∞

· x₁

[c(x₁)]

-∞

-1	1	-1
-1	-1	1
-∞	-1	-∞

· x₁

[a(x₁)]

-∞

0	-∞	2
-1	0	2
-∞	-∞	0

· x₁

the pairs

b^#(x1)	→	a^#(a(x1))	(8)
b^#(x1)	→	a^#(x1)	(9)

could be deleted.

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

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

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair

a^#(b(x1))

→

b^#(c(b(x1)))

(6)

could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.