Certification Problem

Input (TPDB TRS_Innermost/Applicative_AG01_innermost/#4.2)

The rewrite relation of the following TRS is considered.

app(app(app(f,app(g,x)),app(s,0)),y) app(app(app(f,y),y),app(g,x)) (1)
app(g,app(s,x)) app(s,app(g,x)) (2)
app(g,0) 0 (3)
app(app(map,fun),nil) nil (4)
app(app(map,fun),app(app(cons,x),xs)) app(app(cons,app(fun,x)),app(app(map,fun),xs)) (5)
app(app(filter,fun),nil) nil (6)
app(app(filter,fun),app(app(cons,x),xs)) app(app(app(app(filter2,app(fun,x)),fun),x),xs) (7)
app(app(app(app(filter2,true),fun),x),xs) app(app(cons,x),app(app(filter,fun),xs)) (8)
app(app(app(app(filter2,false),fun),x),xs) app(app(filter,fun),xs) (9)
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(app(f,app(g,x)),app(s,0)),y) app#(app(app(f,y),y),app(g,x)) (10)
app#(app(app(f,app(g,x)),app(s,0)),y) app#(app(f,y),y) (11)
app#(app(app(f,app(g,x)),app(s,0)),y) app#(f,y) (12)
app#(g,app(s,x)) app#(s,app(g,x)) (13)
app#(g,app(s,x)) app#(g,x) (14)
app#(app(map,fun),app(app(cons,x),xs)) app#(app(cons,app(fun,x)),app(app(map,fun),xs)) (15)
app#(app(map,fun),app(app(cons,x),xs)) app#(cons,app(fun,x)) (16)
app#(app(map,fun),app(app(cons,x),xs)) app#(fun,x) (17)
app#(app(map,fun),app(app(cons,x),xs)) app#(app(map,fun),xs) (18)
app#(app(filter,fun),app(app(cons,x),xs)) app#(app(app(app(filter2,app(fun,x)),fun),x),xs) (19)
app#(app(filter,fun),app(app(cons,x),xs)) app#(app(app(filter2,app(fun,x)),fun),x) (20)
app#(app(filter,fun),app(app(cons,x),xs)) app#(app(filter2,app(fun,x)),fun) (21)
app#(app(filter,fun),app(app(cons,x),xs)) app#(filter2,app(fun,x)) (22)
app#(app(filter,fun),app(app(cons,x),xs)) app#(fun,x) (23)
app#(app(app(app(filter2,true),fun),x),xs) app#(app(cons,x),app(app(filter,fun),xs)) (24)
app#(app(app(app(filter2,true),fun),x),xs) app#(cons,x) (25)
app#(app(app(app(filter2,true),fun),x),xs) app#(app(filter,fun),xs) (26)
app#(app(app(app(filter2,true),fun),x),xs) app#(filter,fun) (27)
app#(app(app(app(filter2,false),fun),x),xs) app#(app(filter,fun),xs) (28)
app#(app(app(app(filter2,false),fun),x),xs) app#(filter,fun) (29)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.