Certification Problem

Input (TPDB TRS_Standard/Applicative_first_order_05/hydra)

The rewrite relation of the following TRS is considered.

app(f,app(app(cons,nil),y)) → y (1)
app(f,app(app(cons,app(f,app(app(cons,nil),y))),z)) → app(app(app(copy,n),y),z) (2)
app(app(app(copy,0),y),z) → app(f,z) (3)
app(app(app(copy,app(s,x)),y),z) → app(app(app(copy,x),y),app(app(cons,app(f,y)),z)) (4)
app(app(map,fun),nil) → nil (5)
app(app(map,fun),app(app(cons,x),xs)) → app(app(cons,app(fun,x)),app(app(map,fun),xs)) (6)
app(app(filter,fun),nil) → nil (7)
app(app(filter,fun),app(app(cons,x),xs)) → app(app(app(app(filter2,app(fun,x)),fun),x),xs) (8)
app(app(app(app(filter2,true),fun),x),xs) → app(app(cons,x),app(app(filter,fun),xs)) (9)
app(app(app(app(filter2,false),fun),x),xs) → app(app(filter,fun),xs) (10)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Uncurrying

We uncurry the binary symbol app in combination with the following symbol map which also determines the applicative arities of these symbols.

f is mapped to f, f1(x1)
cons is mapped to cons, cons1(x1), cons2(x1, x2)
nil is mapped to nil
copy is mapped to copy, copy1(x1), copy2(x1, x2), copy3(x1, x2, x3)
n is mapped to n
0 is mapped to 0
s is mapped to s, s1(x1)
map is mapped to map, map1(x1), map2(x1, x2)
filter is mapped to filter, filter1(x1), filter3(x1, x2)
filter2 is mapped to filter2, filter21(x1), filter22(x1, x2), filter23(x1, x2, x3), filter24(x1,...,x4)
true is mapped to true
false is mapped to false


There are no uncurry rules.
No rules have to be added for the eta-expansion.

Uncurrying the rules and adding the uncurrying rules yields the new set of rules
f1(cons2(nil,y)) → y (26)
f1(cons2(f1(cons2(nil,y)),z)) → copy3(n,y,z) (27)
copy3(0,y,z) → f1(z) (28)
copy3(s1(x),y,z) → copy3(x,y,cons2(f1(y),z)) (29)
map2(fun,nil) → nil (30)
map2(fun,cons2(x,xs)) → cons2(app(fun,x),map2(fun,xs)) (31)
filter3(fun,nil) → nil (32)
filter3(fun,cons2(x,xs)) → filter24(app(fun,x),fun,x,xs) (33)
filter24(true,fun,x,xs) → cons2(x,filter3(fun,xs)) (34)
filter24(false,fun,x,xs) → filter3(fun,xs) (35)
app(f,y1) → f1(y1) (11)
app(cons,y1) → cons1(y1) (12)
app(cons1(x0),y1) → cons2(x0,y1) (13)
app(copy,y1) → copy1(y1) (14)
app(copy1(x0),y1) → copy2(x0,y1) (15)
app(copy2(x0,x1),y1) → copy3(x0,x1,y1) (16)
app(s,y1) → s1(y1) (17)
app(map,y1) → map1(y1) (18)
app(map1(x0),y1) → map2(x0,y1) (19)
app(filter,y1) → filter1(y1) (20)
app(filter1(x0),y1) → filter3(x0,y1) (21)
app(filter2,y1) → filter21(y1) (22)
app(filter21(x0),y1) → filter22(x0,y1) (23)
app(filter22(x0,x1),y1) → filter23(x0,x1,y1) (24)
app(filter23(x0,x1,x2),y1) → filter24(x0,x1,x2,y1) (25)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
f1#(cons2(f1(cons2(nil,y)),z)) → copy3#(n,y,z) (36)
copy3#(0,y,z) → f1#(z) (37)
copy3#(s1(x),y,z) → copy3#(x,y,cons2(f1(y),z)) (38)
copy3#(s1(x),y,z) → f1#(y) (39)
map2#(fun,cons2(x,xs)) → app#(fun,x) (40)
map2#(fun,cons2(x,xs)) → map2#(fun,xs) (41)
filter3#(fun,cons2(x,xs)) → filter24#(app(fun,x),fun,x,xs) (42)
filter3#(fun,cons2(x,xs)) → app#(fun,x) (43)
filter24#(true,fun,x,xs) → filter3#(fun,xs) (44)
filter24#(false,fun,x,xs) → filter3#(fun,xs) (45)
app#(f,y1) → f1#(y1) (46)
app#(copy2(x0,x1),y1) → copy3#(x0,x1,y1) (47)
app#(map1(x0),y1) → map2#(x0,y1) (48)
app#(filter1(x0),y1) → filter3#(x0,y1) (49)
app#(filter23(x0,x1,x2),y1) → filter24#(x0,x1,x2,y1) (50)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.