Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-289)

The rewrite relation of the following TRS is considered.

a(a(a(a(x1))))	→	a(b(a(b(x1))))	(1)
b(b(a(b(x1))))	→	b(b(b(b(x1))))	(2)
a(a(a(b(x1))))	→	b(a(b(b(x1))))	(3)
b(a(b(b(x1))))	→	b(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 (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers

[b(x₁)]

0	1
0	0

· x₁ +

-∞	-∞
-∞	-∞

[a(x₁)]

0	1
0	1

· x₁ +

-∞	-∞
-∞	-∞

all of the following rules can be deleted.

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

→

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

(1)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

b^#(b(a(b(x1))))	→	b^#(b(x1))	(5)
b^#(b(a(b(x1))))	→	b^#(b(b(x1)))	(6)
b^#(b(a(b(x1))))	→	b^#(b(b(b(x1))))	(7)
a^#(a(a(b(x1))))	→	b^#(b(x1))	(8)
a^#(a(a(b(x1))))	→	a^#(b(b(x1)))	(9)
a^#(a(a(b(x1))))	→	b^#(a(b(b(x1))))	(10)
b^#(a(b(b(x1))))	→	a^#(x1)	(11)
b^#(a(b(b(x1))))	→	b^#(a(x1))	(12)
b^#(a(b(b(x1))))	→	a^#(b(a(x1)))	(13)
b^#(a(b(b(x1))))	→	b^#(a(b(a(x1))))	(14)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

a^#(a(a(b(x1))))	→	b^#(a(b(b(x1))))	(10)
b^#(a(b(b(x1))))	→	b^#(a(b(a(x1))))	(14)
b^#(a(b(b(x1))))	→	b^#(a(x1))	(12)
b^#(a(b(b(x1))))	→	a^#(x1)	(11)

1.1.1.1 Subterm Criterion Processor

We use the projection to multisets

π(a^#)	=	{ 1, 1 }
π(b^#)	=	{ 1, 1 }
π(b)	=	{ 1, 1 }
π(a)	=	{ 1, 1 }

to remove the pairs:

b^#(a(b(b(x1))))	→	b^#(a(x1))	(12)
b^#(a(b(b(x1))))	→	a^#(x1)	(11)

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

→

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

(14)

1.1.1.1.1.1 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	0
0	0	1	0
0	0	0	0
0	1	0	0

· x₁ +

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

[a(x₁)]

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

· x₁ +

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

[b^#(x₁)]

1	0	0	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 rules

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

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

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

→

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

(14)

could be deleted.

1.1.1.1.1.1.1 P is empty

There are no pairs anymore.

The 2^nd component contains the pair

b^#(b(a(b(x1))))	→	b^#(b(b(b(x1))))	(7)
b^#(b(a(b(x1))))	→	b^#(b(b(x1)))	(6)
b^#(b(a(b(x1))))	→	b^#(b(x1))	(5)

1.1.1.2 Subterm Criterion Processor

We use the projection to multisets

π(b^#)	=	{ 1, 1 }
π(b)	=	{ 1, 1 }
π(a)	=	{ 1, 1 }

to remove the pairs:

b^#(b(a(b(x1))))	→	b^#(b(b(x1)))	(6)
b^#(b(a(b(x1))))	→	b^#(b(x1))	(5)

1.1.1.2.1 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)]

0	2
0	0

· x₁ +

0	0
0	0

[a(x₁)]

0	0
0	1

· x₁ +

2	0
1	0

[b^#(x₁)]

2	0
0	0

· x₁ +

1	0
0	0

together with the usable rules

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

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

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

→

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

(7)

could be deleted.

1.1.1.2.1.1 P is empty

There are no pairs anymore.