Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/020)

The rewrite relation of the following TRS is considered.

app(app(plus,0),y)	→	y	(1)
app(app(plus,app(s,x)),y)	→	app(s,app(app(plus,x),y))	(2)
app(app(times,0),y)	→	0	(3)
app(app(times,app(s,x)),y)	→	app(app(plus,app(app(times,x),y)),y)	(4)
app(app(app(comp,f),g),x)	→	app(f,app(g,x))	(5)
app(twice,f)	→	app(app(comp,f),f)	(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.

app^#(app(plus,app(s,x)),y)	→	app^#(plus,x)	(7)
app^#(app(plus,app(s,x)),y)	→	app^#(app(plus,x),y)	(8)
app^#(app(plus,app(s,x)),y)	→	app^#(s,app(app(plus,x),y))	(9)
app^#(app(times,app(s,x)),y)	→	app^#(times,x)	(10)
app^#(app(times,app(s,x)),y)	→	app^#(app(times,x),y)	(11)
app^#(app(times,app(s,x)),y)	→	app^#(plus,app(app(times,x),y))	(12)
app^#(app(times,app(s,x)),y)	→	app^#(app(plus,app(app(times,x),y)),y)	(13)
app^#(app(app(comp,f),g),x)	→	app^#(g,x)	(14)
app^#(app(app(comp,f),g),x)	→	app^#(f,app(g,x))	(15)
app^#(twice,f)	→	app^#(comp,f)	(16)
app^#(twice,f)	→	app^#(app(comp,f),f)	(17)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.

The 1^st component contains the pair

app^#(app(app(comp,f),g),x) → app^#(g,x) (14)

app^#(app(app(comp,f),g),x) → app^#(f,app(g,x)) (15)

1.1.1 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

app^#(app(app(comp,f),g),x) → app^#(g,x) (14)

2 ≥ 2

1 > 1

app^#(app(app(comp,f),g),x) → app^#(f,app(g,x)) (15)

1 > 1

As there is no critical graph in the transitive closure, there are no infinite chains.

The 2^nd component contains the pair

app^#(app(times,app(s,x)),y)

→

app^#(app(times,x),y)

(11)

1.1.2 Reduction Pair Processor with Usable Rules

Using the

prec(app^#)	=	0	stat(app^#)	=	lex
prec(times)	=	0	stat(times)	=	lex
prec(s)	=	0	stat(s)	=	lex
prec(app)	=	1	stat(app)	=	lex

π(app^#)	=	[1]
π(times)	=	[]
π(s)	=	[]
π(app)	=	[2]

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

app^#(app(times,app(s,x)),y)

→

app^#(app(times,x),y)

(11)

could be deleted.

1.1.2.1 P is empty

There are no pairs anymore.

The 3^rd component contains the pair

app^#(app(plus,app(s,x)),y)

→

app^#(app(plus,x),y)

(8)

1.1.3 Reduction Pair Processor with Usable Rules