Certification Problem

Input (TPDB SRS_Standard/Trafo_06/un07)

The rewrite relation of the following TRS is considered.

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

[a(x₁)]

1	0	0
0	0	1
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

[b(x₁)]

1	1	0
0	0	0
0	0	0

· x₁ +

0	0	0
1	0	0
1	0	0

all of the following rules can be deleted.

b(a(b(x1)))

→

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

(3)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

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

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1 over the naturals

[a(x₁)]

1	0	1
0	1	0
1	0	0
0	0	0

· x₁ +

0	0	0	0
0	0	0	0
0	0	0	0
0	0	0	0

[b(x₁)]

0	0	0	0
0	0	0	0
1	0	0	1
1	0	0	1

· x₁ +

0	0	0	0
0	0	0	0
0	0	0	0
1	0	0	0

[b^#(x₁)]

0	0	1	0
0	0	0	0
0	0	0	0
0	0	0	0

· x₁ +

0	0	0	0
0	0	0	0
0	0	0	0
0	0	0	0

together with the usable rules

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

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

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

could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.