Certification Problem

Input (TPDB TRS_Standard/Applicative_first_order_05/#3.48)

The rewrite relation of the following TRS is considered.

app(f,0) true (1)
app(f,1) false (2)
app(f,app(s,x)) app(f,x) (3)
app(app(app(if,true),app(s,x)),app(s,y)) app(s,x) (4)
app(app(app(if,false),app(s,x)),app(s,y)) app(s,y) (5)
app(app(g,x),app(c,y)) app(c,app(app(g,x),y)) (6)
app(app(g,x),app(c,y)) app(app(g,x),app(app(app(if,app(f,x)),app(c,app(app(g,app(s,x)),y))),app(c,y))) (7)
app(app(map,fun),nil) nil (8)
app(app(map,fun),app(app(cons,x),xs)) app(app(cons,app(fun,x)),app(app(map,fun),xs)) (9)
app(app(filter,fun),nil) nil (10)
app(app(filter,fun),app(app(cons,x),xs)) app(app(app(app(filter2,app(fun,x)),fun),x),xs) (11)
app(app(app(app(filter2,true),fun),x),xs) app(app(cons,x),app(app(filter,fun),xs)) (12)
app(app(app(app(filter2,false),fun),x),xs) app(app(filter,fun),xs) (13)

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(g,x),app(c,y)) app#(app(g,x),y) (14)
app#(app(map,fun),app(app(cons,x),xs)) app#(cons,app(fun,x)) (15)
app#(app(filter,fun),app(app(cons,x),xs)) app#(fun,x) (16)
app#(app(filter,fun),app(app(cons,x),xs)) app#(app(app(app(filter2,app(fun,x)),fun),x),xs) (17)
app#(app(filter,fun),app(app(cons,x),xs)) app#(app(filter2,app(fun,x)),fun) (18)
app#(app(g,x),app(c,y)) app#(app(g,app(s,x)),y) (19)
app#(app(g,x),app(c,y)) app#(c,app(app(g,app(s,x)),y)) (20)
app#(app(filter,fun),app(app(cons,x),xs)) app#(filter2,app(fun,x)) (21)
app#(app(app(app(filter2,true),fun),x),xs) app#(app(cons,x),app(app(filter,fun),xs)) (22)
app#(app(g,x),app(c,y)) app#(g,app(s,x)) (23)
app#(app(g,x),app(c,y)) app#(app(if,app(f,x)),app(c,app(app(g,app(s,x)),y))) (24)
app#(app(map,fun),app(app(cons,x),xs)) app#(fun,x) (25)
app#(app(filter,fun),app(app(cons,x),xs)) app#(app(app(filter2,app(fun,x)),fun),x) (26)
app#(app(map,fun),app(app(cons,x),xs)) app#(app(cons,app(fun,x)),app(app(map,fun),xs)) (27)
app#(app(g,x),app(c,y)) app#(f,x) (28)
app#(f,app(s,x)) app#(f,x) (29)
app#(app(app(app(filter2,true),fun),x),xs) app#(filter,fun) (30)
app#(app(g,x),app(c,y)) app#(app(app(if,app(f,x)),app(c,app(app(g,app(s,x)),y))),app(c,y)) (31)
app#(app(g,x),app(c,y)) app#(s,x) (32)
app#(app(g,x),app(c,y)) app#(if,app(f,x)) (33)
app#(app(app(app(filter2,true),fun),x),xs) app#(app(filter,fun),xs) (34)
app#(app(app(app(filter2,false),fun),x),xs) app#(app(filter,fun),xs) (35)
app#(app(g,x),app(c,y)) app#(c,app(app(g,x),y)) (36)
app#(app(map,fun),app(app(cons,x),xs)) app#(app(map,fun),xs) (37)
app#(app(app(app(filter2,true),fun),x),xs) app#(cons,x) (38)
app#(app(app(app(filter2,false),fun),x),xs) app#(filter,fun) (39)
app#(app(g,x),app(c,y)) app#(app(g,x),app(app(app(if,app(f,x)),app(c,app(app(g,app(s,x)),y))),app(c,y))) (40)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.