Certification Problem

Input (TPDB SRS_Standard/Wenzel_16/abaaaa-aaaaabaaabab.srs)

The rewrite relation of the following TRS is considered.

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

→

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

(1)

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(a(a(a(a(x1))))))	→	a^#(b(x1))	(2)
a^#(b(a(a(a(a(x1))))))	→	a^#(b(a(b(x1))))	(3)
a^#(b(a(a(a(a(x1))))))	→	a^#(a(b(a(b(x1)))))	(4)
a^#(b(a(a(a(a(x1))))))	→	a^#(a(a(b(a(b(x1))))))	(5)
a^#(b(a(a(a(a(x1))))))	→	a^#(b(a(a(a(b(a(b(x1))))))))	(6)
a^#(b(a(a(a(a(x1))))))	→	a^#(a(b(a(a(a(b(a(b(x1)))))))))	(7)
a^#(b(a(a(a(a(x1))))))	→	a^#(a(a(b(a(a(a(b(a(b(x1))))))))))	(8)
a^#(b(a(a(a(a(x1))))))	→	a^#(a(a(a(b(a(a(a(b(a(b(x1)))))))))))	(9)
a^#(b(a(a(a(a(x1))))))	→	a^#(a(a(a(a(b(a(a(a(b(a(b(x1))))))))))))	(10)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(b(a(a(a(a(x1))))))	→	a^#(b(a(b(x1))))	(3)
a^#(b(a(a(a(a(x1))))))	→	a^#(b(a(a(a(b(a(b(x1))))))))	(6)
a^#(b(a(a(a(a(x1))))))	→	a^#(b(x1))	(2)

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(x₁)]

=

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

· x₁ +

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

[a(x₁)]

=

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

· x₁ +

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

[a^#(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 rule

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

→

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

(1)

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

a^#(b(a(a(a(a(x1))))))	→	a^#(b(a(b(x1))))	(3)
a^#(b(a(a(a(a(x1))))))	→	a^#(b(a(a(a(b(a(b(x1))))))))	(6)
a^#(b(a(a(a(a(x1))))))	→	a^#(b(x1))	(2)

could be deleted.

1.1.1.1 P is empty

There are no pairs anymore.