Certification Problem

Input (TPDB TRS_Standard/AProVE_04/rta3)

The rewrite relation of the following TRS is considered.

ack(0,y)	→	s(y)	(1)
ack(s(x),0)	→	ack(x,s(0))	(2)
ack(s(x),s(y))	→	ack(x,ack(s(x),y))	(3)
f(s(x),y)	→	f(x,s(x))	(4)
f(x,s(y))	→	f(y,x)	(5)
f(x,y)	→	ack(x,y)	(6)
ack(s(x),y)	→	f(x,x)	(7)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

f^#(x,y)	→	ack^#(x,y)	(8)
ack^#(s(x),s(y))	→	ack^#(s(x),y)	(9)
f^#(s(x),y)	→	f^#(x,s(x))	(10)
ack^#(s(x),s(y))	→	ack^#(x,ack(s(x),y))	(11)
ack^#(s(x),y)	→	f^#(x,x)	(12)
ack^#(s(x),0)	→	ack^#(x,s(0))	(13)
f^#(x,s(y))	→	f^#(y,x)	(14)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(x,s(y))	→	f^#(y,x)	(14)
ack^#(s(x),0)	→	ack^#(x,s(0))	(13)
ack^#(s(x),y)	→	f^#(x,x)	(12)
ack^#(s(x),s(y))	→	ack^#(s(x),y)	(9)
ack^#(s(x),s(y))	→	ack^#(x,ack(s(x),y))	(11)
f^#(s(x),y)	→	f^#(x,s(x))	(10)
f^#(x,y)	→	ack^#(x,y)	(8)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[s(x₁)]	=	x₁ + 2
[ack(x₁, x₂)]	=	max(0)
[f(x₁, x₂)]	=	max(x₁ + 1, 0)
[0]	=	14235
[f^#(x₁, x₂)]	=	max(x₁ + 8857, x₂ + 8856, 0)
[ack^#(x₁, x₂)]	=	max(x₁ + 8857, 0)

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pairs

f^#(x,s(y))	→	f^#(y,x)	(14)
ack^#(s(x),0)	→	ack^#(x,s(0))	(13)
ack^#(s(x),y)	→	f^#(x,x)	(12)
ack^#(s(x),s(y))	→	ack^#(x,ack(s(x),y))	(11)
f^#(s(x),y)	→	f^#(x,s(x))	(10)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

ack^#(s(x),s(y))

→

ack^#(s(x),y)

(9)

1.1.1.1.1 Reduction Pair Processor with Usable Rules

Using the argument filter

in combination with the following Weighted Path Order with the following precedence and status

prec(s)	=	0		status(s)	=	[1]		list-extension(s)	=	Lex
prec(ack)	=	0		status(ack)	=	[1, 2]		list-extension(ack)	=	Lex
prec(f)	=	0		status(f)	=	[1, 2]		list-extension(f)	=	Lex
prec(0)	=	0		status(0)	=	[]		list-extension(0)	=	Lex
prec(f^#)	=	0		status(f^#)	=	[1, 2]		list-extension(f^#)	=	Lex
prec(ack^#)	=	0		status(ack^#)	=	[1, 2]		list-extension(ack^#)	=	Lex

and the following Max-polynomial interpretation

[s(x₁)]	=	x₁ + 0
[ack(x₁, x₂)]	=	max(x₁ + 0, x₂ + 0, 0)
[f(x₁, x₂)]	=	max(x₁ + 0, x₂ + 0, 0)
[0]	=	0
[f^#(x₁, x₂)]	=	max(x₁ + 0, x₂ + 0, 0)
[ack^#(x₁, x₂)]	=	max(x₁ + 0, x₂ + 0, 0)

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair

ack^#(s(x),s(y))

→

ack^#(s(x),y)

(9)

could be deleted.

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.