Certification Problem

Input (TPDB TRS_Standard/AProVE_07/thiemann40)

The rewrite relation of the following TRS is considered.

nonZero(0)	→	false	(1)
nonZero(s(x))	→	true	(2)
p(0)	→	0	(3)
p(s(x))	→	x	(4)
id_inc(x)	→	x	(5)
id_inc(x)	→	s(x)	(6)
random(x)	→	rand(x,0)	(7)
rand(x,y)	→	if(nonZero(x),x,y)	(8)
if(false,x,y)	→	y	(9)
if(true,x,y)	→	rand(p(x),id_inc(y))	(10)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

random^#(x)	→	rand^#(x,0)	(11)
rand^#(x,y)	→	if^#(nonZero(x),x,y)	(12)
rand^#(x,y)	→	nonZero^#(x)	(13)
if^#(true,x,y)	→	rand^#(p(x),id_inc(y))	(14)
if^#(true,x,y)	→	p^#(x)	(15)
if^#(true,x,y)	→	id_inc^#(y)	(16)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

if^#(true,x,y)	→	rand^#(p(x),id_inc(y))	(14)
rand^#(x,y)	→	if^#(nonZero(x),x,y)	(12)

1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the rationals with delta = 1/4

[if^#(x₁, x₂, x₃)]	=	0 + 1/4 · x₁ + 2 · x₂ + 0 · x₃
[true]	=	1
[rand^#(x₁, x₂)]	=	1/4 + 4 · x₁ + 0 · x₂
[p(x₁)]	=	0 + 1/4 · x₁
[id_inc(x₁)]	=	0 + 0 · x₁
[nonZero(x₁)]	=	0 + 4 · x₁
[0]	=	0
[s(x₁)]	=	1/2 + 4 · x₁
[false]	=	0

together with the usable rules

p(0)	→	0	(3)
p(s(x))	→	x	(4)
nonZero(0)	→	false	(1)
nonZero(s(x))	→	true	(2)

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

rand^#(x,y)

→

if^#(nonZero(x),x,y)

(12)

could be deleted.

1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[p(x₁)]	=	1 · x₁
[0]	=	0
[s(x₁)]	=	1 · x₁
[id_inc(x₁)]	=	1 · x₁
[true]	=	0
[rand^#(x₁, x₂)]	=	1 · x₁ + 1 · x₂
[if^#(x₁, x₂, x₃)]	=	1 · x₁ + 1 · x₂ + 1 · x₃

together with the usable rules

p(0)	→	0	(3)
p(s(x))	→	x	(4)
id_inc(x)	→	x	(5)
id_inc(x)	→	s(x)	(6)

(w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[p(x₁)]	=	2 + 2 · x₁
[0]	=	2
[s(x₁)]	=	2 + 1 · x₁
[id_inc(x₁)]	=	2 + 1 · x₁
[if^#(x₁, x₂, x₃)]	=	2 + 2 · x₁ + 2 · x₂ + 1 · x₃
[true]	=	2
[rand^#(x₁, x₂)]	=	2 + 1 · x₁ + 1 · x₂

together with the usable rules

p(0)	→	0	(3)
p(s(x))	→	x	(4)
id_inc(x)	→	x	(5)
id_inc(x)	→	s(x)	(6)

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

if^#(true,x,y)

→

rand^#(p(x),id_inc(y))

(14)

and the rules

p(0)	→	0	(3)
p(s(x))	→	x	(4)
id_inc(x)	→	x	(5)

could be deleted.

1.1.1.1.1.1 P is empty

There are no pairs anymore.