Certification Problem

Input (TPDB TRS_Standard/Zantema_05/z03)

The rewrite relation of the following TRS is considered.

a(f,a(g,a(f,x))) a(f,a(g,a(g,a(f,x)))) (1)
a(g,a(f,a(g,x))) a(g,a(f,a(f,a(g,x)))) (2)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Uncurrying

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

f is mapped to f, f1(x1)
g is mapped to g, g1(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
f1(g1(f1(x))) f1(g1(g1(f1(x)))) (5)
g1(f1(g1(x))) g1(f1(f1(g1(x)))) (6)
a(f,y1) f1(y1) (3)
a(g,y1) g1(y1) (4)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[f1(x1)] = 1 · x1
[g1(x1)] = 1 · x1
[a(x1, x2)] = 1 · x1 + 2 · x2
[f] = 0
[g] = 1
all of the following rules can be deleted.
a(g,y1) g1(y1) (4)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[f1(x1)] = 1 · x1
[g1(x1)] = 1 · x1
[a(x1, x2)] = 1 + 2 · x1 + 2 · x2
[f] = 0
all of the following rules can be deleted.
a(f,y1) f1(y1) (3)

1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
f1#(g1(f1(x))) f1#(g1(g1(f1(x)))) (7)
f1#(g1(f1(x))) g1#(g1(f1(x))) (8)
g1#(f1(g1(x))) g1#(f1(f1(g1(x)))) (9)
g1#(f1(g1(x))) f1#(f1(g1(x))) (10)

1.1.1.1.1 Reduction Pair Processor

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(f1#) = 1 weight(f1#) = 1
prec(g1) = 0 weight(g1) = 1
prec(f1) = 2 weight(f1) = 1
in combination with the following argument filter

π(f1#) = []
π(g1#) = 1
π(g1) = []
π(f1) = []

the pairs
f1#(g1(f1(x))) g1#(g1(f1(x))) (8)
g1#(f1(g1(x))) f1#(f1(g1(x))) (10)
could be deleted.

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.