Certification Problem

Input (TPDB SRS_Standard/Bouchare_06/07)

The rewrite relation of the following TRS is considered.

b(a(x1))	→	b(b(x1))	(1)
b(a(b(x1)))	→	b(a(a(x1)))	(2)
a(a(a(x1)))	→	a(b(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.

b^#(a(x1))	→	b^#(x1)	(4)
b^#(a(x1))	→	b^#(b(x1))	(5)
b^#(a(b(x1)))	→	a^#(x1)	(6)
b^#(a(b(x1)))	→	a^#(a(x1))	(7)
b^#(a(b(x1)))	→	b^#(a(a(x1)))	(8)
a^#(a(a(x1)))	→	b^#(x1)	(9)
a^#(a(a(x1)))	→	b^#(b(x1))	(10)
a^#(a(a(x1)))	→	a^#(b(b(x1)))	(11)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(a(a(x1)))	→	b^#(x1)	(9)
b^#(a(b(x1)))	→	b^#(a(a(x1)))	(8)
b^#(a(b(x1)))	→	a^#(a(x1))	(7)
b^#(a(b(x1)))	→	a^#(x1)	(6)
b^#(a(x1))	→	b^#(x1)	(4)

1.1.1 Subterm Criterion Processor

We use the projection to multisets

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

to remove the pairs:

a^#(a(a(x1)))	→	b^#(x1)	(9)
b^#(a(b(x1)))	→	a^#(a(x1))	(7)
b^#(a(b(x1)))	→	a^#(x1)	(6)
b^#(a(x1))	→	b^#(x1)	(4)

1.1.1.1 Reduction Pair Processor with Usable Rules

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

[b(x₁)]

=

0	0	0
0	0	0
1	0	0

· x₁ +

0	0	0
0	0	0
1	0	0

[a(x₁)]

=

0	0	1
0	0	1
1	0	0

· x₁ +

1	0	0
0	0	0
0	0	0

[b^#(x₁)]

=

0	1	0
0	0	0
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

together with the usable rules

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

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

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

→

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

(8)

could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.