Certification Problem

Input (TPDB TRS_Standard/AProVE_07/wiehe11)

The rewrite relation of the following TRS is considered.

minus(x,0) x (1)
minus(s(x),s(y)) minus(x,y) (2)
quot(0,s(y)) 0 (3)
quot(s(x),s(y)) s(quot(minus(x,y),s(y))) (4)
plus(0,y) y (5)
plus(s(x),y) s(plus(x,y)) (6)
minus(minus(x,y),z) minus(x,plus(y,z)) (7)
app(nil,k) k (8)
app(l,nil) l (9)
app(cons(x,l),k) cons(x,app(l,k)) (10)
sum(cons(x,nil)) cons(x,nil) (11)
sum(cons(x,cons(y,l))) sum(cons(plus(x,y),l)) (12)
sum(app(l,cons(x,cons(y,k)))) sum(app(l,sum(cons(x,cons(y,k))))) (13)
plus(s(x),s(y)) s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))))) (14)
plus(s(x),x) plus(if(gt(x,x),id(x),id(x)),s(x)) (15)
plus(zero,y) y (16)
plus(id(x),s(y)) s(plus(x,if(gt(s(y),y),y,s(y)))) (17)
id(x) x (18)
if(true,x,y) x (19)
if(false,x,y) y (20)
not(x) if(x,false,true) (21)
gt(s(x),zero) true (22)
gt(zero,y) false (23)
gt(s(x),s(y)) gt(x,y) (24)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
minus#(s(x),s(y)) minus#(x,y) (25)
quot#(s(x),s(y)) quot#(minus(x,y),s(y)) (26)
quot#(s(x),s(y)) minus#(x,y) (27)
plus#(s(x),y) plus#(x,y) (28)
minus#(minus(x,y),z) minus#(x,plus(y,z)) (29)
minus#(minus(x,y),z) plus#(y,z) (30)
app#(cons(x,l),k) app#(l,k) (31)
sum#(cons(x,cons(y,l))) sum#(cons(plus(x,y),l)) (32)
sum#(cons(x,cons(y,l))) plus#(x,y) (33)
sum#(app(l,cons(x,cons(y,k)))) sum#(app(l,sum(cons(x,cons(y,k))))) (34)
sum#(app(l,cons(x,cons(y,k)))) app#(l,sum(cons(x,cons(y,k)))) (35)
sum#(app(l,cons(x,cons(y,k)))) sum#(cons(x,cons(y,k))) (36)
plus#(s(x),s(y)) plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))) (37)
plus#(s(x),s(y)) if#(gt(x,y),x,y) (38)
plus#(s(x),s(y)) gt#(x,y) (39)
plus#(s(x),s(y)) if#(not(gt(x,y)),id(x),id(y)) (40)
plus#(s(x),s(y)) not#(gt(x,y)) (41)
plus#(s(x),s(y)) id#(x) (42)
plus#(s(x),s(y)) id#(y) (43)
plus#(s(x),x) plus#(if(gt(x,x),id(x),id(x)),s(x)) (44)
plus#(s(x),x) if#(gt(x,x),id(x),id(x)) (45)
plus#(s(x),x) gt#(x,x) (46)
plus#(s(x),x) id#(x) (47)
plus#(id(x),s(y)) plus#(x,if(gt(s(y),y),y,s(y))) (48)
plus#(id(x),s(y)) if#(gt(s(y),y),y,s(y)) (49)
plus#(id(x),s(y)) gt#(s(y),y) (50)
not#(x) if#(x,false,true) (51)
gt#(s(x),s(y)) gt#(x,y) (52)

1.1 Dependency Graph Processor

The dependency pairs are split into 7 components.