Certification Problem

Input (TPDB TRS_Standard/Applicative_first_order_05/18)

The rewrite relation of the following TRS is considered.

app(app(*,x),app(app(+,y),z)) app(app(+,app(app(*,x),y)),app(app(*,x),z)) (1)
app(app(map,f),nil) nil (2)
app(app(map,f),app(app(cons,x),xs)) app(app(cons,app(f,x)),app(app(map,f),xs)) (3)
app(app(filter,f),nil) nil (4)
app(app(filter,f),app(app(cons,x),xs)) app(app(app(app(filter2,app(f,x)),f),x),xs) (5)
app(app(app(app(filter2,true),f),x),xs) app(app(cons,x),app(app(filter,f),xs)) (6)
app(app(app(app(filter2,false),f),x),xs) app(app(filter,f),xs) (7)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
app#(app(*,x),app(app(+,y),z)) app#(app(*,x),z) (8)
app#(app(*,x),app(app(+,y),z)) app#(app(*,x),y) (9)
app#(app(*,x),app(app(+,y),z)) app#(+,app(app(*,x),y)) (10)
app#(app(*,x),app(app(+,y),z)) app#(app(+,app(app(*,x),y)),app(app(*,x),z)) (11)
app#(app(map,f),app(app(cons,x),xs)) app#(app(map,f),xs) (12)
app#(app(map,f),app(app(cons,x),xs)) app#(f,x) (13)
app#(app(map,f),app(app(cons,x),xs)) app#(cons,app(f,x)) (14)
app#(app(map,f),app(app(cons,x),xs)) app#(app(cons,app(f,x)),app(app(map,f),xs)) (15)
app#(app(filter,f),app(app(cons,x),xs)) app#(f,x) (16)
app#(app(filter,f),app(app(cons,x),xs)) app#(filter2,app(f,x)) (17)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(filter2,app(f,x)),f) (18)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(app(filter2,app(f,x)),f),x) (19)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(app(app(filter2,app(f,x)),f),x),xs) (20)
app#(app(app(app(filter2,true),f),x),xs) app#(filter,f) (21)
app#(app(app(app(filter2,true),f),x),xs) app#(app(filter,f),xs) (22)
app#(app(app(app(filter2,true),f),x),xs) app#(cons,x) (23)
app#(app(app(app(filter2,true),f),x),xs) app#(app(cons,x),app(app(filter,f),xs)) (24)
app#(app(app(app(filter2,false),f),x),xs) app#(filter,f) (25)
app#(app(app(app(filter2,false),f),x),xs) app#(app(filter,f),xs) (26)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.