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 ttt2 @ 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)	→	nonZero^#(x)	(12)
rand^#(x,y)	→	if^#(nonZero(x),x,y)	(13)
if^#(true,x,y)	→	id_inc^#(y)	(14)
if^#(true,x,y)	→	p^#(x)	(15)
if^#(true,x,y)	→	rand^#(p(x),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))	(16)
rand^#(x,y)	→	if^#(nonZero(x),x,y)	(13)

1.1.1 Reduction Pair Processor with Usable Rules

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

[true]	=	1
[if^#(x₁, x₂, x₃)]	=	0 · x₁ + -7 · x₂ + -∞ · x₃ + -16
[nonZero(x₁)]	=	0 · x₁ + -4
[s(x₁)]	=	9 · x₁ + 1
[p(x₁)]	=	-8 · x₁ + 0
[0]	=	0
[rand^#(x₁, x₂)]	=	0 · x₁ + -∞ · x₂ + 0
[id_inc(x₁)]	=	6 · x₁ + 1
[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

if^#(true,x,y)

→

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

(16)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.