YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

a(d(x0))	→	d(b(x0))
a(x0)	→	b(b(b(x0)))
d(x0)	→	x0
a(x0)	→	x0
b(d(b(x0)))	→	a(d(x0))
b(c(x0))	→	c(d(d(x0)))
a(c(x0))	→	b(b(c(d(x0))))

Proof

1 Rule Removal

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

[a(x₁)]

2	0
2	0

· x₁ +

-∞	-∞
-∞	-∞

[c(x₁)]

0	-∞
0	-∞

· x₁ +

-∞	-∞
-∞	-∞

[d(x₁)]

0	-∞
0	2

· x₁ +

-∞	-∞
-∞	-∞

[b(x₁)]

0	0
0	0

· x₁ +

-∞	-∞
-∞	-∞

the rules

a(d(x0))	→	d(b(x0))
a(x0)	→	b(b(b(x0)))
d(x0)	→	x0
a(x0)	→	x0
b(d(b(x0)))	→	a(d(x0))
b(c(x0))	→	c(d(d(x0)))

remain.

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

a^#(d(x0))	→	b^#(x0)
a^#(d(x0))	→	d^#(b(x0))
a^#(x0)	→	b^#(x0)
a^#(x0)	→	b^#(b(x0))
a^#(x0)	→	b^#(b(b(x0)))
b^#(d(b(x0)))	→	d^#(x0)
b^#(d(b(x0)))	→	a^#(d(x0))
b^#(c(x0))	→	d^#(x0)
b^#(c(x0))	→	d^#(d(x0))

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pairs

b^#(d(b(x0)))	→	a^#(d(x0))
a^#(d(x0))	→	b^#(x0)
a^#(x0)	→	b^#(x0)
a^#(x0)	→	b^#(b(x0))
a^#(x0)	→	b^#(b(b(x0)))

1.1.1.1 Reduction Pair Processor with Usable Rules

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

[a^#(x₁)]	=	0 · x₁ + -∞
[a(x₁)]	=	0 · x₁ + -∞
[c(x₁)]	=	-∞ · x₁ + 6
[b^#(x₁)]	=	0 · x₁ + -∞
[d(x₁)]	=	8 · x₁ + -∞
[b(x₁)]	=	0 · x₁ + -∞

together with the usable rules

a(d(x0))	→	d(b(x0))
a(x0)	→	b(b(b(x0)))
d(x0)	→	x0
a(x0)	→	x0
b(d(b(x0)))	→	a(d(x0))
b(c(x0))	→	c(d(d(x0)))

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

b^#(d(b(x0)))	→	a^#(d(x0))
a^#(x0)	→	b^#(x0)
a^#(x0)	→	b^#(b(x0))
a^#(x0)	→	b^#(b(b(x0)))

remain.

1.1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

[a^#(x₁)]	=	1 · x₁ + 2
[a(x₁)]	=	1 · x₁ + 5
[c(x₁)]	=	0 · x₁ + 1
[b^#(x₁)]	=	1 · x₁ + 0
[d(x₁)]	=	4 · x₁ + 8
[b(x₁)]	=	1 · x₁ + 1

the pair

a^#(x0)

→

b^#(b(b(x0)))

remains.

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.