Certification Problem

Input (TPDB SRS_Standard/Wenzel_16/aabaababaab-aababaababaaba.srs)

The rewrite relation of the following TRS is considered.

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

→

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

(1)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS

a_a(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(x1)))))))))))	→	a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(a_a(x1))))))))))))))	(2)
a_a(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_b(x1)))))))))))	→	a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(a_b(x1))))))))))))))	(3)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

a_a^#(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(x1)))))))))))	→	a_a^#(x1)	(4)
a_a^#(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(x1)))))))))))	→	a_a^#(a_b(b_a(a_a(x1))))	(5)
a_a^#(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(x1)))))))))))	→	a_a^#(a_b(b_a(a_b(b_a(a_a(a_b(b_a(a_a(x1)))))))))	(6)
a_a^#(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(x1)))))))))))	→	a_a^#(a_b(b_a(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(a_a(x1))))))))))))))	(7)
a_a^#(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_b(x1)))))))))))	→	a_a^#(a_b(b_a(a_b(x1))))	(8)
a_a^#(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_b(x1)))))))))))	→	a_a^#(a_b(b_a(a_b(b_a(a_a(a_b(b_a(a_b(x1)))))))))	(9)
a_a^#(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_b(x1)))))))))))	→	a_a^#(a_b(b_a(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(a_b(x1))))))))))))))	(10)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a_a^#(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(x1)))))))))))	→	a_a^#(a_b(b_a(a_a(x1))))	(5)
a_a^#(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(x1)))))))))))	→	a_a^#(x1)	(4)

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

[b_a(x₁)]

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

· x₁ +

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

[a_b(x₁)]

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

· x₁ +

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

[b_b(x₁)]

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

· x₁ +

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

[a_a^#(x₁)]

0	1	0	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

[a_a(x₁)]

1	1	0
0	0	0
0	1	1
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

a_a(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(x1)))))))))))	→	a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(a_a(x1))))))))))))))	(2)
a_a(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_b(x1)))))))))))	→	a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(a_b(x1))))))))))))))	(3)

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

a_a^#(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(x1)))))))))))	→	a_a^#(a_b(b_a(a_a(x1))))	(5)
a_a^#(a_b(b_a(a_a(a_b(b_a(a_b(b_a(a_a(a_b(b_a(x1)))))))))))	→	a_a^#(x1)	(4)

could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.