Certification Problem

Input (TPDB TRS_Standard/Applicative_first_order_05/#3.16)

The rewrite relation of the following TRS is considered.

app(app(times,x),0) 0 (1)
app(app(times,x),app(s,y)) app(app(plus,app(app(times,x),y)),x) (2)
app(app(plus,x),0) x (3)
app(app(plus,0),x) x (4)
app(app(plus,x),app(s,y)) app(s,app(app(plus,x),y)) (5)
app(app(plus,app(s,x)),y) app(s,app(app(plus,x),y)) (6)
app(app(map,f),nil) nil (7)
app(app(map,f),app(app(cons,x),xs)) app(app(cons,app(f,x)),app(app(map,f),xs)) (8)
app(app(filter,f),nil) nil (9)
app(app(filter,f),app(app(cons,x),xs)) app(app(app(app(filter2,app(f,x)),f),x),xs) (10)
app(app(app(app(filter2,true),f),x),xs) app(app(cons,x),app(app(filter,f),xs)) (11)
app(app(app(app(filter2,false),f),x),xs) app(app(filter,f),xs) (12)

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(times,x),app(s,y)) app#(app(times,x),y) (13)
app#(app(times,x),app(s,y)) app#(plus,app(app(times,x),y)) (14)
app#(app(times,x),app(s,y)) app#(app(plus,app(app(times,x),y)),x) (15)
app#(app(plus,x),app(s,y)) app#(app(plus,x),y) (16)
app#(app(plus,x),app(s,y)) app#(s,app(app(plus,x),y)) (17)
app#(app(plus,app(s,x)),y) app#(plus,x) (18)
app#(app(plus,app(s,x)),y) app#(app(plus,x),y) (19)
app#(app(plus,app(s,x)),y) app#(s,app(app(plus,x),y)) (20)
app#(app(map,f),app(app(cons,x),xs)) app#(app(map,f),xs) (21)
app#(app(map,f),app(app(cons,x),xs)) app#(f,x) (22)
app#(app(map,f),app(app(cons,x),xs)) app#(cons,app(f,x)) (23)
app#(app(map,f),app(app(cons,x),xs)) app#(app(cons,app(f,x)),app(app(map,f),xs)) (24)
app#(app(filter,f),app(app(cons,x),xs)) app#(f,x) (25)
app#(app(filter,f),app(app(cons,x),xs)) app#(filter2,app(f,x)) (26)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(filter2,app(f,x)),f) (27)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(app(filter2,app(f,x)),f),x) (28)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(app(app(filter2,app(f,x)),f),x),xs) (29)
app#(app(app(app(filter2,true),f),x),xs) app#(filter,f) (30)
app#(app(app(app(filter2,true),f),x),xs) app#(app(filter,f),xs) (31)
app#(app(app(app(filter2,true),f),x),xs) app#(cons,x) (32)
app#(app(app(app(filter2,true),f),x),xs) app#(app(cons,x),app(app(filter,f),xs)) (33)
app#(app(app(app(filter2,false),f),x),xs) app#(filter,f) (34)
app#(app(app(app(filter2,false),f),x),xs) app#(app(filter,f),xs) (35)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.