Certification Problem

Input (TPDB TRS_Innermost/Applicative_AG01_innermost/#4.15)

The rewrite relation of the following TRS is considered.

app(app(app(app(f,0),1),app(app(g,x),y)),z) app(app(app(app(f,app(app(g,x),y)),app(app(g,x),y)),app(app(g,x),y)),app(h,x)) (1)
app(app(g,0),1) 0 (2)
app(app(g,0),1) 1 (3)
app(h,app(app(g,x),y)) app(h,x) (4)
app(app(map,fun),nil) nil (5)
app(app(map,fun),app(app(cons,x),xs)) app(app(cons,app(fun,x)),app(app(map,fun),xs)) (6)
app(app(filter,fun),nil) nil (7)
app(app(filter,fun),app(app(cons,x),xs)) app(app(app(app(filter2,app(fun,x)),fun),x),xs) (8)
app(app(app(app(filter2,true),fun),x),xs) app(app(cons,x),app(app(filter,fun),xs)) (9)
app(app(app(app(filter2,false),fun),x),xs) app(app(filter,fun),xs) (10)
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(app(f,0),1),app(app(g,x),y)),z) app#(app(app(app(f,app(app(g,x),y)),app(app(g,x),y)),app(app(g,x),y)),app(h,x)) (11)
app#(app(app(app(f,0),1),app(app(g,x),y)),z) app#(app(app(f,app(app(g,x),y)),app(app(g,x),y)),app(app(g,x),y)) (12)
app#(app(app(app(f,0),1),app(app(g,x),y)),z) app#(app(f,app(app(g,x),y)),app(app(g,x),y)) (13)
app#(app(app(app(f,0),1),app(app(g,x),y)),z) app#(f,app(app(g,x),y)) (14)
app#(app(app(app(f,0),1),app(app(g,x),y)),z) app#(h,x) (15)
app#(h,app(app(g,x),y)) app#(h,x) (16)
app#(app(map,fun),app(app(cons,x),xs)) app#(app(cons,app(fun,x)),app(app(map,fun),xs)) (17)
app#(app(map,fun),app(app(cons,x),xs)) app#(cons,app(fun,x)) (18)
app#(app(map,fun),app(app(cons,x),xs)) app#(fun,x) (19)
app#(app(map,fun),app(app(cons,x),xs)) app#(app(map,fun),xs) (20)
app#(app(filter,fun),app(app(cons,x),xs)) app#(app(app(app(filter2,app(fun,x)),fun),x),xs) (21)
app#(app(filter,fun),app(app(cons,x),xs)) app#(app(app(filter2,app(fun,x)),fun),x) (22)
app#(app(filter,fun),app(app(cons,x),xs)) app#(app(filter2,app(fun,x)),fun) (23)
app#(app(filter,fun),app(app(cons,x),xs)) app#(filter2,app(fun,x)) (24)
app#(app(filter,fun),app(app(cons,x),xs)) app#(fun,x) (25)
app#(app(app(app(filter2,true),fun),x),xs) app#(app(cons,x),app(app(filter,fun),xs)) (26)
app#(app(app(app(filter2,true),fun),x),xs) app#(cons,x) (27)
app#(app(app(app(filter2,true),fun),x),xs) app#(app(filter,fun),xs) (28)
app#(app(app(app(filter2,true),fun),x),xs) app#(filter,fun) (29)
app#(app(app(app(filter2,false),fun),x),xs) app#(app(filter,fun),xs) (30)
app#(app(app(app(filter2,false),fun),x),xs) app#(filter,fun) (31)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.