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.

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

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)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.