Certification Problem

Input (TPDB TRS_Standard/Applicative_first_order_05/17)

The rewrite relation of the following TRS is considered.

app(app(.,1),x) x (1)
app(app(.,x),1) x (2)
app(app(.,app(i,x)),x) 1 (3)
app(app(.,x),app(i,x)) 1 (4)
app(app(.,app(i,y)),app(app(.,y),z)) z (5)
app(app(.,y),app(app(.,app(i,y)),z)) z (6)
app(app(.,app(app(.,x),y)),z) app(app(.,x),app(app(.,y),z)) (7)
app(i,1) 1 (8)
app(i,app(i,x)) x (9)
app(i,app(app(.,x),y)) app(app(.,app(i,y)),app(i,x)) (10)
app(app(map,f),nil) nil (11)
app(app(map,f),app(app(cons,x),xs)) app(app(cons,app(f,x)),app(app(map,f),xs)) (12)
app(app(filter,f),nil) nil (13)
app(app(filter,f),app(app(cons,x),xs)) app(app(app(app(filter2,app(f,x)),f),x),xs) (14)
app(app(app(app(filter2,true),f),x),xs) app(app(cons,x),app(app(filter,f),xs)) (15)
app(app(app(app(filter2,false),f),x),xs) app(app(filter,f),xs) (16)

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(.,app(app(.,x),y)),z) app#(.,y) (17)
app#(app(.,app(app(.,x),y)),z) app#(app(.,y),z) (18)
app#(app(.,app(app(.,x),y)),z) app#(app(.,x),app(app(.,y),z)) (19)
app#(i,app(app(.,x),y)) app#(i,x) (20)
app#(i,app(app(.,x),y)) app#(i,y) (21)
app#(i,app(app(.,x),y)) app#(.,app(i,y)) (22)
app#(i,app(app(.,x),y)) app#(app(.,app(i,y)),app(i,x)) (23)
app#(app(map,f),app(app(cons,x),xs)) app#(app(map,f),xs) (24)
app#(app(map,f),app(app(cons,x),xs)) app#(f,x) (25)
app#(app(map,f),app(app(cons,x),xs)) app#(cons,app(f,x)) (26)
app#(app(map,f),app(app(cons,x),xs)) app#(app(cons,app(f,x)),app(app(map,f),xs)) (27)
app#(app(filter,f),app(app(cons,x),xs)) app#(f,x) (28)
app#(app(filter,f),app(app(cons,x),xs)) app#(filter2,app(f,x)) (29)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(filter2,app(f,x)),f) (30)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(app(filter2,app(f,x)),f),x) (31)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(app(app(filter2,app(f,x)),f),x),xs) (32)
app#(app(app(app(filter2,true),f),x),xs) app#(filter,f) (33)
app#(app(app(app(filter2,true),f),x),xs) app#(app(filter,f),xs) (34)
app#(app(app(app(filter2,true),f),x),xs) app#(cons,x) (35)
app#(app(app(app(filter2,true),f),x),xs) app#(app(cons,x),app(app(filter,f),xs)) (36)
app#(app(app(app(filter2,false),f),x),xs) app#(filter,f) (37)
app#(app(app(app(filter2,false),f),x),xs) app#(app(filter,f),xs) (38)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.