Certification Problem

Input (TPDB TRS_Standard/Applicative_first_order_05/12)

The rewrite relation of the following TRS is considered.

app(not,app(not,x)) x (1)
app(not,app(app(or,x),y)) app(app(and,app(not,x)),app(not,y)) (2)
app(not,app(app(and,x),y)) app(app(or,app(not,x)),app(not,y)) (3)
app(app(and,x),app(app(or,y),z)) app(app(or,app(app(and,x),y)),app(app(and,x),z)) (4)
app(app(and,app(app(or,y),z)),x) app(app(or,app(app(and,x),y)),app(app(and,x),z)) (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 ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
app#(not,app(app(or,x),y)) app#(not,y) (12)
app#(not,app(app(or,x),y)) app#(not,x) (13)
app#(not,app(app(or,x),y)) app#(and,app(not,x)) (14)
app#(not,app(app(or,x),y)) app#(app(and,app(not,x)),app(not,y)) (15)
app#(not,app(app(and,x),y)) app#(not,y) (16)
app#(not,app(app(and,x),y)) app#(not,x) (17)
app#(not,app(app(and,x),y)) app#(or,app(not,x)) (18)
app#(not,app(app(and,x),y)) app#(app(or,app(not,x)),app(not,y)) (19)
app#(app(and,x),app(app(or,y),z)) app#(app(and,x),z) (20)
app#(app(and,x),app(app(or,y),z)) app#(app(and,x),y) (21)
app#(app(and,x),app(app(or,y),z)) app#(or,app(app(and,x),y)) (22)
app#(app(and,x),app(app(or,y),z)) app#(app(or,app(app(and,x),y)),app(app(and,x),z)) (23)
app#(app(and,app(app(or,y),z)),x) app#(app(and,x),z) (24)
app#(app(and,app(app(or,y),z)),x) app#(and,x) (25)
app#(app(and,app(app(or,y),z)),x) app#(app(and,x),y) (26)
app#(app(and,app(app(or,y),z)),x) app#(or,app(app(and,x),y)) (27)
app#(app(and,app(app(or,y),z)),x) app#(app(or,app(app(and,x),y)),app(app(and,x),z)) (28)
app#(app(map,f),app(app(cons,x),xs)) app#(app(map,f),xs) (29)
app#(app(map,f),app(app(cons,x),xs)) app#(f,x) (30)
app#(app(map,f),app(app(cons,x),xs)) app#(cons,app(f,x)) (31)
app#(app(map,f),app(app(cons,x),xs)) app#(app(cons,app(f,x)),app(app(map,f),xs)) (32)
app#(app(filter,f),app(app(cons,x),xs)) app#(f,x) (33)
app#(app(filter,f),app(app(cons,x),xs)) app#(filter2,app(f,x)) (34)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(filter2,app(f,x)),f) (35)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(app(filter2,app(f,x)),f),x) (36)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(app(app(filter2,app(f,x)),f),x),xs) (37)
app#(app(app(app(filter2,true),f),x),xs) app#(filter,f) (38)
app#(app(app(app(filter2,true),f),x),xs) app#(app(filter,f),xs) (39)
app#(app(app(app(filter2,true),f),x),xs) app#(cons,x) (40)
app#(app(app(app(filter2,true),f),x),xs) app#(app(cons,x),app(app(filter,f),xs)) (41)
app#(app(app(app(filter2,false),f),x),xs) app#(filter,f) (42)
app#(app(app(app(filter2,false),f),x),xs) app#(app(filter,f),xs) (43)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.