Certification Problem

Input (TPDB SRS_Standard/Secret_07_SRS/x02)

The rewrite relation of the following TRS is considered.

a(b(x1))	→	c(d(x1))	(1)
d(d(x1))	→	b(e(x1))	(2)
b(x1)	→	d(c(x1))	(3)
d(x1)	→	x1	(4)
e(c(x1))	→	d(a(x1))	(5)
a(x1)	→	e(d(x1))	(6)

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(x1))	→	d^#(x1)	(7)
d^#(d(x1))	→	e^#(x1)	(8)
d^#(d(x1))	→	b^#(e(x1))	(9)
b^#(x1)	→	d^#(c(x1))	(10)
e^#(c(x1))	→	a^#(x1)	(11)
e^#(c(x1))	→	d^#(a(x1))	(12)
a^#(x1)	→	d^#(x1)	(13)
a^#(x1)	→	e^#(d(x1))	(14)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

e^#(c(x1))	→	d^#(a(x1))	(12)
d^#(d(x1))	→	e^#(x1)	(8)
e^#(c(x1))	→	a^#(x1)	(11)
a^#(x1)	→	e^#(d(x1))	(14)
a^#(x1)	→	d^#(x1)	(13)
a^#(b(x1))	→	d^#(x1)	(7)

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 arctic semiring over the integers

[e(x₁)]

0	-∞
0	-∞

· x₁ +

-∞	-∞
0	-∞

[a(x₁)]

0	0
0	0

· x₁ +

1	-∞
1	-∞

[c(x₁)]

1	1
1	0

· x₁ +

2	-∞
2	-∞

[a^#(x₁)]

0	0
-∞	-∞

· x₁ +

1	-∞
-∞	-∞

[b(x₁)]

1	1
2	2

· x₁ +

2	-∞
3	-∞

[e^#(x₁)]

0	-∞
-∞	-∞

· x₁ +

0	-∞
-∞	-∞

[d^#(x₁)]

0	0
-∞	-∞

· x₁ +

0	-∞
-∞	-∞

[d(x₁)]

0	0
1	1

· x₁ +

1	-∞
2	-∞

together with the usable rules

a(b(x1))	→	c(d(x1))	(1)
d(d(x1))	→	b(e(x1))	(2)
b(x1)	→	d(c(x1))	(3)
d(x1)	→	x1	(4)
e(c(x1))	→	d(a(x1))	(5)
a(x1)	→	e(d(x1))	(6)

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

e^#(c(x1))	→	d^#(a(x1))	(12)
d^#(d(x1))	→	e^#(x1)	(8)
e^#(c(x1))	→	a^#(x1)	(11)
a^#(b(x1))	→	d^#(x1)	(7)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.