Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-14)

The rewrite relation of the following TRS is considered.

b(f(b(x,z)),y)	→	f(f(f(b(z,b(y,z)))))	(1)
c(f(f(c(x,a,z))),a,y)	→	b(y,f(b(a,z)))	(2)
b(b(c(b(a,a),a,z),f(a)),y)	→	z	(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^#(f(b(x,z)),y)	→	b^#(y,z)	(4)
b^#(f(b(x,z)),y)	→	b^#(z,b(y,z))	(5)
c^#(f(f(c(x,a,z))),a,y)	→	b^#(a,z)	(6)
c^#(f(f(c(x,a,z))),a,y)	→	b^#(y,f(b(a,z)))	(7)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

b^#(f(b(x,z)),y)	→	b^#(z,b(y,z))	(5)
b^#(f(b(x,z)),y)	→	b^#(y,z)	(4)

1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals

[b^#(x₁, x₂)]

1	0
0	0

· x₁ +

2	0
0	0

· x₂ +

0	0
0	0

[f(x₁)]

0	1
0	0

· x₁ +

0	0
1	0

[c(x₁, x₂, x₃)]

0	0
0	0

· x₁ +

0	0
0	0

· x₂ +

1	0
0	2

· x₃ +

0	0
0	0

[b(x₁, x₂)]

1	0
0	1

· x₁ +

0	0
2	2

· x₂ +

0	0
2	0

[a]

2	0
0	0

together with the usable rules

b(f(b(x,z)),y)	→	f(f(f(b(z,b(y,z)))))	(1)
b(b(c(b(a,a),a,z),f(a)),y)	→	z	(3)

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

b^#(f(b(x,z)),y)	→	b^#(z,b(y,z))	(5)
b^#(f(b(x,z)),y)	→	b^#(y,z)	(4)

could be deleted.

1.1.1.1 P is empty

There are no pairs anymore.