Certification Problem

Input (TPDB TRS_Standard/Applicative_first_order_05/08)

The rewrite relation of the following TRS is considered.

app(D,t) 1 (1)
app(D,constant) 0 (2)
app(D,app(app(+,x),y)) app(app(+,app(D,x)),app(D,y)) (3)
app(D,app(app(*,x),y)) app(app(+,app(app(*,y),app(D,x))),app(app(*,x),app(D,y))) (4)
app(D,app(app(-,x),y)) app(app(-,app(D,x)),app(D,y)) (5)
app(app(map,f),nil) nil (6)
app(app(map,f),app(app(cons,x),xs)) app(app(cons,app(f,x)),app(app(map,f),xs)) (7)
app(app(filter,f),nil) nil (8)
app(app(filter,f),app(app(cons,x),xs)) app(app(app(app(filter2,app(f,x)),f),x),xs) (9)
app(app(app(app(filter2,true),f),x),xs) app(app(cons,x),app(app(filter,f),xs)) (10)
app(app(app(app(filter2,false),f),x),xs) app(app(filter,f),xs) (11)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
app#(app(filter,f),app(app(cons,x),xs)) app#(app(app(filter2,app(f,x)),f),x) (12)
app#(D,app(app(*,x),y)) app#(app(*,x),app(D,y)) (13)
app#(app(app(app(filter2,false),f),x),xs) app#(app(filter,f),xs) (14)
app#(app(app(app(filter2,true),f),x),xs) app#(cons,x) (15)
app#(app(map,f),app(app(cons,x),xs)) app#(cons,app(f,x)) (16)
app#(app(map,f),app(app(cons,x),xs)) app#(app(map,f),xs) (17)
app#(D,app(app(*,x),y)) app#(app(+,app(app(*,y),app(D,x))),app(app(*,x),app(D,y))) (18)
app#(D,app(app(+,x),y)) app#(D,y) (19)
app#(app(app(app(filter2,true),f),x),xs) app#(app(cons,x),app(app(filter,f),xs)) (20)
app#(app(app(app(filter2,true),f),x),xs) app#(app(filter,f),xs) (21)
app#(D,app(app(*,x),y)) app#(D,y) (22)
app#(D,app(app(*,x),y)) app#(+,app(app(*,y),app(D,x))) (23)
app#(D,app(app(+,x),y)) app#(app(+,app(D,x)),app(D,y)) (24)
app#(app(app(app(filter2,false),f),x),xs) app#(filter,f) (25)
app#(app(map,f),app(app(cons,x),xs)) app#(f,x) (26)
app#(app(filter,f),app(app(cons,x),xs)) app#(f,x) (27)
app#(D,app(app(*,x),y)) app#(D,x) (28)
app#(D,app(app(+,x),y)) app#(D,x) (29)
app#(D,app(app(*,x),y)) app#(*,y) (30)
app#(D,app(app(-,x),y)) app#(-,app(D,x)) (31)
app#(D,app(app(-,x),y)) app#(D,y) (32)
app#(D,app(app(*,x),y)) app#(app(*,y),app(D,x)) (33)
app#(D,app(app(+,x),y)) app#(+,app(D,x)) (34)
app#(D,app(app(-,x),y)) app#(app(-,app(D,x)),app(D,y)) (35)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(filter2,app(f,x)),f) (36)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(app(app(filter2,app(f,x)),f),x),xs) (37)
app#(app(map,f),app(app(cons,x),xs)) app#(app(cons,app(f,x)),app(app(map,f),xs)) (38)
app#(app(app(app(filter2,true),f),x),xs) app#(filter,f) (39)
app#(app(filter,f),app(app(cons,x),xs)) app#(filter2,app(f,x)) (40)
app#(D,app(app(-,x),y)) app#(D,x) (41)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.