Certification Problem

Input (TPDB SRS_Standard/Trafo_06/hom03)

The rewrite relation of the following TRS is considered.

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

→

a(a(a(a(b(c(a(b(c(a(b(c(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(c(a(b(c(a(a(a(x1)))))))))	→	a^#(b(c(x1)))	(2)
a^#(b(c(a(b(c(a(a(a(x1)))))))))	→	a^#(b(c(a(b(c(x1))))))	(3)
a^#(b(c(a(b(c(a(a(a(x1)))))))))	→	a^#(b(c(a(b(c(a(b(c(x1)))))))))	(4)
a^#(b(c(a(b(c(a(a(a(x1)))))))))	→	a^#(a(b(c(a(b(c(a(b(c(x1))))))))))	(5)
a^#(b(c(a(b(c(a(a(a(x1)))))))))	→	a^#(a(a(b(c(a(b(c(a(b(c(x1)))))))))))	(6)
a^#(b(c(a(b(c(a(a(a(x1)))))))))	→	a^#(a(a(a(b(c(a(b(c(a(b(c(x1))))))))))))	(7)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(b(c(a(b(c(a(a(a(x1)))))))))	→	a^#(b(c(a(b(c(a(b(c(x1)))))))))	(4)
a^#(b(c(a(b(c(a(a(a(x1)))))))))	→	a^#(b(c(a(b(c(x1))))))	(3)
a^#(b(c(a(b(c(a(a(a(x1)))))))))	→	a^#(b(c(x1)))	(2)

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 rationals with delta = 1/64

[c(x₁)]

=

0	0	1
0	0	0
0	1	0

· x₁ +

0	0	0
0	0	0
0	0	0

[a^#(x₁)]

=

0	0	1
0	0	0
0	0	0

· x₁ +

1	0	0
0	0	0
0	0	0

[a(x₁)]

=

0	1	0
0	0	1
1	0	0

· x₁ +

0	0	0
0	0	0
1	0	0

[b(x₁)]

=

0	0	1
1	0	0
0	0	1

· x₁ +

1	0	0
1	0	0
0	0	0

together with the usable rule

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

→

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

(1)

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

a^#(b(c(a(b(c(a(a(a(x1)))))))))	→	a^#(b(c(a(b(c(a(b(c(x1)))))))))	(4)
a^#(b(c(a(b(c(a(a(a(x1)))))))))	→	a^#(b(c(a(b(c(x1))))))	(3)
a^#(b(c(a(b(c(a(a(a(x1)))))))))	→	a^#(b(c(x1)))	(2)

could be deleted.

1.1.1.1 P is empty

There are no pairs anymore.