Certification Problem

Input (TPDB TRS_Innermost/Applicative_AG01_innermost/#4.10)

The rewrite relation of the following TRS is considered.

app(app(f,x),app(g,x))	→	app(app(f,1),app(g,x))	(1)
app(g,1)	→	app(g,0)	(2)
app(app(map,fun),nil)	→	nil	(3)
app(app(map,fun),app(app(cons,x),xs))	→	app(app(cons,app(fun,x)),app(app(map,fun),xs))	(4)
app(app(filter,fun),nil)	→	nil	(5)
app(app(filter,fun),app(app(cons,x),xs))	→	app(app(app(app(filter2,app(fun,x)),fun),x),xs)	(6)
app(app(app(app(filter2,true),fun),x),xs)	→	app(app(cons,x),app(app(filter,fun),xs))	(7)
app(app(app(app(filter2,false),fun),x),xs)	→	app(app(filter,fun),xs)	(8)

The evaluation strategy is innermost.

Prove or disprove termination.

Yes.

The following set of initial dependency pairs has been identified.

app^#(app(f,x),app(g,x))	→	app^#(app(f,1),app(g,x))	(9)
app^#(app(f,x),app(g,x))	→	app^#(f,1)	(10)
app^#(g,1)	→	app^#(g,0)	(11)
app^#(app(map,fun),app(app(cons,x),xs))	→	app^#(app(cons,app(fun,x)),app(app(map,fun),xs))	(12)
app^#(app(map,fun),app(app(cons,x),xs))	→	app^#(cons,app(fun,x))	(13)
app^#(app(map,fun),app(app(cons,x),xs))	→	app^#(fun,x)	(14)
app^#(app(map,fun),app(app(cons,x),xs))	→	app^#(app(map,fun),xs)	(15)
app^#(app(filter,fun),app(app(cons,x),xs))	→	app^#(app(app(app(filter2,app(fun,x)),fun),x),xs)	(16)
app^#(app(filter,fun),app(app(cons,x),xs))	→	app^#(app(app(filter2,app(fun,x)),fun),x)	(17)
app^#(app(filter,fun),app(app(cons,x),xs))	→	app^#(app(filter2,app(fun,x)),fun)	(18)
app^#(app(filter,fun),app(app(cons,x),xs))	→	app^#(filter2,app(fun,x))	(19)
app^#(app(filter,fun),app(app(cons,x),xs))	→	app^#(fun,x)	(20)
app^#(app(app(app(filter2,true),fun),x),xs)	→	app^#(app(cons,x),app(app(filter,fun),xs))	(21)
app^#(app(app(app(filter2,true),fun),x),xs)	→	app^#(cons,x)	(22)
app^#(app(app(app(filter2,true),fun),x),xs)	→	app^#(app(filter,fun),xs)	(23)
app^#(app(app(app(filter2,true),fun),x),xs)	→	app^#(filter,fun)	(24)
app^#(app(app(app(filter2,false),fun),x),xs)	→	app^#(app(filter,fun),xs)	(25)
app^#(app(app(app(filter2,false),fun),x),xs)	→	app^#(filter,fun)	(26)

The dependency pairs are split into 1 component.

The 1^st component contains the pair

app^#(app(map,fun),app(app(cons,x),xs))	→	app^#(app(map,fun),xs)	(15)
app^#(app(map,fun),app(app(cons,x),xs))	→	app^#(fun,x)	(14)
app^#(app(filter,fun),app(app(cons,x),xs))	→	app^#(fun,x)	(20)
app^#(app(app(app(filter2,true),fun),x),xs)	→	app^#(app(filter,fun),xs)	(23)
app^#(app(app(app(filter2,false),fun),x),xs)	→	app^#(app(filter,fun),xs)	(25)

We restrict the rewrite rules to the following usable rules of the DP problem.

There are no rules.

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

app^#(app(filter,fun),app(app(cons,x),xs))	→	app^#(fun,x)	(20)

1	>	1
2	>	2
app^#(app(map,fun),app(app(cons,x),xs))	→	app^#(fun,x)	(14)

1	>	1
2	>	2
app^#(app(map,fun),app(app(cons,x),xs))	→	app^#(app(map,fun),xs)	(15)

1	≥	1
2	>	2
app^#(app(app(app(filter2,true),fun),x),xs)	→	app^#(app(filter,fun),xs)	(23)

2	≥	2
app^#(app(app(app(filter2,false),fun),x),xs)	→	app^#(app(filter,fun),xs)	(25)

2	≥	2

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