Certification Problem

Input (TPDB TRS_Standard/Zantema_05/jw28)

The rewrite relation of the following TRS is considered.

f(f(a,x),a) → f(a,f(f(a,f(a,a)),x)) (1)

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), a2(x1, x2)


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
a2(x,a) → a1(a2(a1(a),x)) (4)
f(a,y1) → a1(y1) (2)
f(a1(x0),y1) → a2(x0,y1) (3)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a2(x1, x2)] = 1 + 2 · x1 + 2 · x2
[a] = 1
[a1(x1)] = 1 · x1
[f(x1, x2)] = 2 + 2 · x1 + 2 · x2
all of the following rules can be deleted.
f(a,y1) → a1(y1) (2)
f(a1(x0),y1) → a2(x0,y1) (3)

1.1.1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a2#(x,a) → a2#(a1(a),x) (5)

1.1.1.1.1 Usable Rules Processor

We restrict the rewrite rules to the following usable rules of the DP problem.

There are no rules.

1.1.1.1.1.1 Innermost Lhss Removal Processor

We restrict the innermost strategy to the following left hand sides.

There are no lhss.

1.1.1.1.1.1.1 Reduction Pair Processor

Using the
prec(a2#) = 1 stat(a2#) = mul
prec(a) = 1 stat(a) = mul
prec(a1) = 0 stat(a1) = mul

π(a2#) = [1,2]
π(a) = []
π(a1) = []

the pair
a2#(x,a) → a2#(a1(a),x) (5)
could be deleted.

1.1.1.1.1.1.1.1 P is empty

There are no pairs anymore.