Certification Problem

Input (TPDB TRS_Innermost/Applicative_AG01_innermost/#4.19)

The rewrite relation of the following TRS is considered.

app(app(app(f,x),app(c,x)),app(c,y)) app(app(app(f,y),y),app(app(app(f,y),x),y)) (1)
app(app(app(f,app(s,x)),y),z) app(app(app(f,x),app(s,app(c,y))),app(c,z)) (2)
app(app(app(f,app(c,x)),x),y) app(c,y) (3)
app(app(g,x),y) x (4)
app(app(g,x),y) y (5)
app(app(map,fun),nil) nil (6)
app(app(map,fun),app(app(cons,x),xs)) app(app(cons,app(fun,x)),app(app(map,fun),xs)) (7)
app(app(filter,fun),nil) nil (8)
app(app(filter,fun),app(app(cons,x),xs)) app(app(app(app(filter2,app(fun,x)),fun),x),xs) (9)
app(app(app(app(filter2,true),fun),x),xs) app(app(cons,x),app(app(filter,fun),xs)) (10)
app(app(app(app(filter2,false),fun),x),xs) app(app(filter,fun),xs) (11)
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,x),app(c,x)),app(c,y)) app#(app(app(f,y),y),app(app(app(f,y),x),y)) (12)
app#(app(app(f,x),app(c,x)),app(c,y)) app#(app(f,y),y) (13)
app#(app(app(f,x),app(c,x)),app(c,y)) app#(f,y) (14)
app#(app(app(f,x),app(c,x)),app(c,y)) app#(app(app(f,y),x),y) (15)
app#(app(app(f,x),app(c,x)),app(c,y)) app#(app(f,y),x) (16)
app#(app(app(f,app(s,x)),y),z) app#(app(app(f,x),app(s,app(c,y))),app(c,z)) (17)
app#(app(app(f,app(s,x)),y),z) app#(app(f,x),app(s,app(c,y))) (18)
app#(app(app(f,app(s,x)),y),z) app#(f,x) (19)
app#(app(app(f,app(s,x)),y),z) app#(s,app(c,y)) (20)
app#(app(app(f,app(s,x)),y),z) app#(c,y) (21)
app#(app(app(f,app(s,x)),y),z) app#(c,z) (22)
app#(app(app(f,app(c,x)),x),y) app#(c,y) (23)
app#(app(map,fun),app(app(cons,x),xs)) app#(app(cons,app(fun,x)),app(app(map,fun),xs)) (24)
app#(app(map,fun),app(app(cons,x),xs)) app#(cons,app(fun,x)) (25)
app#(app(map,fun),app(app(cons,x),xs)) app#(fun,x) (26)
app#(app(map,fun),app(app(cons,x),xs)) app#(app(map,fun),xs) (27)
app#(app(filter,fun),app(app(cons,x),xs)) app#(app(app(app(filter2,app(fun,x)),fun),x),xs) (28)
app#(app(filter,fun),app(app(cons,x),xs)) app#(app(app(filter2,app(fun,x)),fun),x) (29)
app#(app(filter,fun),app(app(cons,x),xs)) app#(app(filter2,app(fun,x)),fun) (30)
app#(app(filter,fun),app(app(cons,x),xs)) app#(filter2,app(fun,x)) (31)
app#(app(filter,fun),app(app(cons,x),xs)) app#(fun,x) (32)
app#(app(app(app(filter2,true),fun),x),xs) app#(app(cons,x),app(app(filter,fun),xs)) (33)
app#(app(app(app(filter2,true),fun),x),xs) app#(cons,x) (34)
app#(app(app(app(filter2,true),fun),x),xs) app#(app(filter,fun),xs) (35)
app#(app(app(app(filter2,true),fun),x),xs) app#(filter,fun) (36)
app#(app(app(app(filter2,false),fun),x),xs) app#(app(filter,fun),xs) (37)
app#(app(app(app(filter2,false),fun),x),xs) app#(filter,fun) (38)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.