Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/028)

The rewrite relation of the following TRS is considered.

app(app(app(consif,true),x),ys)	→	app(app(cons,x),ys)	(1)
app(app(app(consif,false),x),ys)	→	ys	(2)
app(app(filter,f),nil)	→	nil	(3)
app(app(filter,f),app(app(cons,x),xs))	→	app(app(app(consif,app(f,x)),x),app(app(filter,f),xs))	(4)

Prove or disprove termination.

Yes.

The following set of initial dependency pairs has been identified.

app^#(app(app(consif,true),x),ys)	→	app^#(cons,x)	(5)
app^#(app(filter,f),app(app(cons,x),xs))	→	app^#(f,x)	(6)
app^#(app(filter,f),app(app(cons,x),xs))	→	app^#(app(filter,f),xs)	(7)
app^#(app(filter,f),app(app(cons,x),xs))	→	app^#(consif,app(f,x))	(8)
app^#(app(filter,f),app(app(cons,x),xs))	→	app^#(app(app(consif,app(f,x)),x),app(app(filter,f),xs))	(9)
app^#(app(app(consif,true),x),ys)	→	app^#(app(cons,x),ys)	(10)
app^#(app(filter,f),app(app(cons,x),xs))	→	app^#(app(consif,app(f,x)),x)	(11)

The dependency pairs are split into 1 component.

The 1^st component contains the pair

app^#(app(filter,f),app(app(cons,x),xs))	→	app^#(f,x)	(6)
app^#(app(filter,f),app(app(cons,x),xs))	→	app^#(app(filter,f),xs)	(7)

Using the Max-polynomial interpretation

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

app^#(app(filter,f),app(app(cons,x),xs))	→	app^#(f,x)	(6)
app^#(app(filter,f),app(app(cons,x),xs))	→	app^#(app(filter,f),xs)	(7)

could be deleted.

The dependency pairs are split into 0 components.