Certification Problem

Input (TPDB TRS_Innermost/Applicative_AG01_innermost/#4.5)

The rewrite relation of the following TRS is considered.

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

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)	(13)
app^#(app(map,fun),app(app(cons,x),xs))	→	app^#(fun,x)	(12)
app^#(app(filter,fun),app(app(cons,x),xs))	→	app^#(fun,x)	(18)
app^#(app(app(app(filter2,true),fun),x),xs)	→	app^#(app(filter,fun),xs)	(21)
app^#(app(app(app(filter2,false),fun),x),xs)	→	app^#(app(filter,fun),xs)	(23)

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

There are no rules.

We restrict the innermost strategy to the following left hand sides.

app(app(map,x0),nil)

app(app(map,x0),app(app(cons,x1),x2))

app(app(filter,x0),nil)

app(app(filter,x0),app(app(cons,x1),x2))

app(app(app(app(filter2,true),x0),x1),x2)

app(app(app(app(filter2,false),x0),x1),x2)

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)	(18)

1	>	1
2	>	2
app^#(app(map,fun),app(app(cons,x),xs))	→	app^#(fun,x)	(12)

1	>	1
2	>	2
app^#(app(map,fun),app(app(cons,x),xs))	→	app^#(app(map,fun),xs)	(13)

1	≥	1
2	>	2
app^#(app(app(app(filter2,true),fun),x),xs)	→	app^#(app(filter,fun),xs)	(21)

2	≥	2
app^#(app(app(app(filter2,false),fun),x),xs)	→	app^#(app(filter,fun),xs)	(23)

2	≥	2

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