Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-470)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals

[b(x₁)]

1	1	0
0	0	1
0	0	0

· x₁ +

0	0	0
0	0	0
1	0	0

[a(x₁)]

1	0	0
0	0	0
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

all of the following rules can be deleted.

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

→

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

(4)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

a^#(b(a(a(x1))))	→	b^#(a(x1))	(5)
a^#(b(a(a(x1))))	→	b^#(b(a(x1)))	(6)
a^#(b(a(a(x1))))	→	a^#(b(b(a(x1))))	(7)
a^#(b(a(b(x1))))	→	a^#(x1)	(8)
a^#(b(a(b(x1))))	→	b^#(a(x1))	(9)
a^#(b(a(b(x1))))	→	b^#(b(a(x1)))	(10)
a^#(b(a(b(x1))))	→	a^#(b(b(a(x1))))	(11)
b^#(b(a(a(x1))))	→	a^#(b(a(a(x1))))	(12)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

b^#(b(a(a(x1))))	→	a^#(b(a(a(x1))))	(12)
a^#(b(a(b(x1))))	→	a^#(b(b(a(x1))))	(11)
a^#(b(a(b(x1))))	→	b^#(b(a(x1)))	(10)
a^#(b(a(b(x1))))	→	a^#(x1)	(8)
a^#(b(a(a(x1))))	→	a^#(b(b(a(x1))))	(7)
a^#(b(a(a(x1))))	→	b^#(b(a(x1)))	(6)

1.1.1.1 Subterm Criterion Processor

We use the projection to multisets

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

to remove the pairs:

a^#(b(a(b(x1))))	→	b^#(b(a(x1)))	(10)
a^#(b(a(b(x1))))	→	a^#(x1)	(8)
a^#(b(a(a(x1))))	→	b^#(b(a(x1)))	(6)

1.1.1.1.1 Dependency Graph Processor