Certification Problem

Input (TPDB SRS_Standard/Zantema_06/13)

The rewrite relation of the following TRS is considered.

a(a(a(b(b(x1)))))	→	b(b(b(x1)))	(1)
b(a(a(a(b(x1)))))	→	a(a(a(b(a(a(a(x1)))))))	(2)
a(a(a(x1)))	→	a(a(x1))	(3)
b(b(x1))	→	a(b(a(x1)))	(4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers

[a(x₁)]	=	2 · x₁ + -∞
[b(x₁)]	=	6 · x₁ + -∞

all of the following rules can be deleted.

a(a(a(x1)))	→	a(a(x1))	(3)
b(b(x1))	→	a(b(a(x1)))	(4)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

a^#(a(a(b(b(x1)))))	→	b^#(b(b(x1)))	(5)
b^#(a(a(a(b(x1)))))	→	a^#(x1)	(6)
b^#(a(a(a(b(x1)))))	→	a^#(a(x1))	(7)
b^#(a(a(a(b(x1)))))	→	a^#(a(a(x1)))	(8)
b^#(a(a(a(b(x1)))))	→	b^#(a(a(a(x1))))	(9)
b^#(a(a(a(b(x1)))))	→	a^#(b(a(a(a(x1)))))	(10)
b^#(a(a(a(b(x1)))))	→	a^#(a(b(a(a(a(x1))))))	(11)
b^#(a(a(a(b(x1)))))	→	a^#(a(a(b(a(a(a(x1)))))))	(12)

1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[a(x₁)]	=	1 · x₁ + 1
[b^#(x₁)]	=	2 · x₁ + 4
[b(x₁)]	=	1 · x₁ + 3
[a^#(x₁)]	=	2 · x₁ + 0

together with the usable rules

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

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

b^#(a(a(a(b(x1)))))	→	a^#(x1)	(6)
b^#(a(a(a(b(x1)))))	→	a^#(a(x1))	(7)
b^#(a(a(a(b(x1)))))	→	a^#(a(a(x1)))	(8)
b^#(a(a(a(b(x1)))))	→	b^#(a(a(a(x1))))	(9)
b^#(a(a(a(b(x1)))))	→	a^#(b(a(a(a(x1)))))	(10)
b^#(a(a(a(b(x1)))))	→	a^#(a(b(a(a(a(x1))))))	(11)

could be deleted.

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the arctic semiring over the integers

[a(x₁)]

0	0	0
0	-∞	0
-∞	-∞	0

· x₁ +

0	-∞	-∞
0	-∞	-∞
0	-∞	-∞

[b^#(x₁)]

0	-∞	0
-∞	-∞	-∞
-∞	-∞	-∞

· x₁ +

0	-∞	-∞
-∞	-∞	-∞
-∞	-∞	-∞

[b(x₁)]

0	-∞	0
0	0	1
0	-∞	0

· x₁ +

0	-∞	-∞
1	-∞	-∞
0	-∞	-∞

[a^#(x₁)]

0	0	0
-∞	-∞	-∞
-∞	-∞	-∞

· x₁ +

0	-∞	-∞
-∞	-∞	-∞
-∞	-∞	-∞

together with the usable rules

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

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

a^#(a(a(b(b(x1)))))

→

b^#(b(b(x1)))

(5)

could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.