Certification Problem

Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-354)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

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

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

1.1.1 Subterm Criterion Processor

We use the projection to multisets

π(a^#)	=	{ 1 }
π(c)	=	{ 1 }
π(b)	=	{ 1, 1 }
π(a)	=	{ 1, 1 }

to remove the pairs:

a^#(b(x1))

→

a^#(x1)

(4)

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers

[b(x₁)]

-∞	0
1	-∞

· x₁ +

0	-∞
2	-∞

[a^#(x₁)]

-∞	1
-∞	-∞

· x₁ +

0	-∞
-∞	-∞

[a(x₁)]

-∞	0
1	-∞

· x₁ +

1	-∞
2	-∞

[c(x₁)]

1	-∞
-∞	-1

· x₁ +

2	-∞
-4	-∞

together with the usable rules

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

(w.r.t. the implicit argument filter of the reduction pair), the pair

a^#(b(x1))

→

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

(6)

could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.