Certification Problem

Input (TPDB TRS_Standard/Zantema_05/z07)

The rewrite relation of the following TRS is considered.

f(a,x) f(b,f(c,x)) (1)
f(a,f(b,x)) f(b,f(a,x)) (2)
f(d,f(c,x)) f(d,f(a,x)) (3)
f(a,f(c,x)) f(c,f(a,x)) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Uncurrying

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

a is mapped to a, a1(x1)
b is mapped to b, b1(x1)
c is mapped to c, c1(x1)
d is mapped to d, d1(x1)


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
a1(x) b1(c1(x)) (9)
a1(b1(x)) b1(a1(x)) (10)
d1(c1(x)) d1(a1(x)) (11)
a1(c1(x)) c1(a1(x)) (12)
f(a,y1) a1(y1) (5)
f(b,y1) b1(y1) (6)
f(c,y1) c1(y1) (7)
f(d,y1) d1(y1) (8)

1.1 Rule Removal

Using the
prec(d1) = 0 stat(d1) = mul
prec(f) = 1 stat(f) = mul
prec(a) = 2 stat(a) = mul
prec(b) = 3 stat(b) = mul
prec(c) = 4 stat(c) = mul
prec(d) = 5 stat(d) = mul

π(a1) = 1
π(b1) = 1
π(c1) = 1
π(d1) = [1]
π(f) = [1,2]
π(a) = []
π(b) = []
π(c) = []
π(d) = []

all of the following rules can be deleted.
f(a,y1) a1(y1) (5)
f(b,y1) b1(y1) (6)
f(c,y1) c1(y1) (7)
f(d,y1) d1(y1) (8)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a1#(b1(x)) a1#(x) (13)
d1#(c1(x)) d1#(a1(x)) (14)
d1#(c1(x)) a1#(x) (15)
a1#(c1(x)) a1#(x) (16)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.