Certification Problem

Input (TPDB SRS_Standard/Trafo_06/dup08)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

b^#(b(a(a(x1))))	→	a^#(b(b(b(b(x1)))))	(4)
b^#(b(a(a(x1))))	→	b^#(x1)	(5)
a^#(a(a(a(b(b(x1))))))	→	b^#(b(a(a(b(b(x1))))))	(6)
b^#(b(c(c(a(a(x1))))))	→	a^#(a(b(b(x1))))	(7)
b^#(b(c(c(a(a(x1))))))	→	b^#(b(x1))	(8)
b^#(b(c(c(a(a(x1))))))	→	b^#(x1)	(9)
a^#(a(a(a(b(b(x1))))))	→	b^#(a(a(b(b(x1)))))	(10)
b^#(b(c(c(a(a(x1))))))	→	a^#(a(a(a(b(b(x1))))))	(11)
b^#(b(c(c(a(a(x1))))))	→	a^#(b(b(x1)))	(12)
b^#(b(a(a(x1))))	→	b^#(b(x1))	(13)
b^#(b(a(a(x1))))	→	b^#(b(b(b(x1))))	(14)
b^#(b(a(a(x1))))	→	a^#(a(b(b(b(b(x1))))))	(15)
b^#(b(c(c(a(a(x1))))))	→	a^#(a(a(b(b(x1)))))	(16)
b^#(b(a(a(x1))))	→	b^#(b(b(x1)))	(17)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

b^#(b(a(a(x1))))	→	b^#(b(b(x1)))	(17)
a^#(a(a(a(b(b(x1))))))	→	b^#(a(a(b(b(x1)))))	(10)
b^#(b(c(c(a(a(x1))))))	→	b^#(x1)	(9)
b^#(b(c(c(a(a(x1))))))	→	a^#(a(a(b(b(x1)))))	(16)
b^#(b(a(a(x1))))	→	a^#(a(b(b(b(b(x1))))))	(15)
b^#(b(a(a(x1))))	→	b^#(b(b(b(x1))))	(14)
b^#(b(c(c(a(a(x1))))))	→	b^#(b(x1))	(8)
b^#(b(c(c(a(a(x1))))))	→	a^#(a(b(b(x1))))	(7)
a^#(a(a(a(b(b(x1))))))	→	b^#(b(a(a(b(b(x1))))))	(6)
b^#(b(a(a(x1))))	→	b^#(x1)	(5)
b^#(b(a(a(x1))))	→	b^#(b(x1))	(13)
b^#(b(c(c(a(a(x1))))))	→	a^#(b(b(x1)))	(12)
b^#(b(c(c(a(a(x1))))))	→	a^#(a(a(a(b(b(x1))))))	(11)
b^#(b(a(a(x1))))	→	a^#(b(b(b(b(x1)))))	(4)

1.1.1 Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals

[a(x₁)]

0	0
0	1

· x₁ +

[b(x₁)]

0	0
0	1

· x₁ +

[c(x₁)]

0	0
1	1

· x₁ +

[a^#(x₁)]

0	1
0	0

· x₁ +

[b^#(x₁)]

0	1
0	0

· x₁ +

together with the usable rules

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

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

b^#(b(c(c(a(a(x1))))))	→	b^#(x1)	(9)
b^#(b(c(c(a(a(x1))))))	→	a^#(a(a(b(b(x1)))))	(16)
b^#(b(c(c(a(a(x1))))))	→	b^#(b(x1))	(8)
b^#(b(c(c(a(a(x1))))))	→	a^#(a(b(b(x1))))	(7)
b^#(b(c(c(a(a(x1))))))	→	a^#(b(b(x1)))	(12)
b^#(b(c(c(a(a(x1))))))	→	a^#(a(a(a(b(b(x1))))))	(11)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(a(a(a(b(b(x1))))))	→	b^#(a(a(b(b(x1)))))	(10)
a^#(a(a(a(b(b(x1))))))	→	b^#(b(a(a(b(b(x1))))))	(6)
b^#(b(a(a(x1))))	→	b^#(x1)	(5)
b^#(b(a(a(x1))))	→	b^#(b(x1))	(13)
b^#(b(a(a(x1))))	→	b^#(b(b(x1)))	(17)
b^#(b(a(a(x1))))	→	b^#(b(b(b(x1))))	(14)
b^#(b(a(a(x1))))	→	a^#(b(b(b(b(x1)))))	(4)
b^#(b(a(a(x1))))	→	a^#(a(b(b(b(b(x1))))))	(15)

1.1.1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[a(x₁)]	=	x₁ + 1
[b(x₁)]	=	x₁ + 0
[c(x₁)]	=	10451
[a^#(x₁)]	=	x₁ + 0
[b^#(x₁)]	=	x₁ + 0

together with the usable rules

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

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

a^#(a(a(a(b(b(x1))))))	→	b^#(a(a(b(b(x1)))))	(10)
a^#(a(a(a(b(b(x1))))))	→	b^#(b(a(a(b(b(x1))))))	(6)
b^#(b(a(a(x1))))	→	b^#(x1)	(5)
b^#(b(a(a(x1))))	→	b^#(b(x1))	(13)
b^#(b(a(a(x1))))	→	b^#(b(b(x1)))	(17)
b^#(b(a(a(x1))))	→	b^#(b(b(b(x1))))	(14)
b^#(b(a(a(x1))))	→	a^#(b(b(b(b(x1)))))	(4)
b^#(b(a(a(x1))))	→	a^#(a(b(b(b(b(x1))))))	(15)

could be deleted.

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.