Certification Problem

Input (TPDB TRS_Standard/Waldmann_06/jwteparla2)

The rewrite relation of the following TRS is considered.

f(f(f(a,f(a,a)),a),x) f(x,f(x,a)) (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), a3(x1, x2, x3)


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
a3(a1(a),a,x) f(x,f(x,a)) (5)
f(a,y1) a1(y1) (2)
f(a1(x0),y1) a2(x0,y1) (3)
f(a2(x0,x1),y1) a3(x0,x1,y1) (4)

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 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a3#(a1(a),a,x) f#(x,f(x,a)) (6)
a3#(a1(a),a,x) f#(x,a) (7)
f#(a2(x0,x1),y1) a3#(x0,x1,y1) (8)

1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[f#(x1, x2)] = -1 + x1 + x2
[f(x1, x2)] = 2
[a] = 0
[a1(x1)] = 0
[a2(x1, x2)] = 2
[a3(x1, x2, x3)] = 2
[a3#(x1, x2, x3)] = 1 + x3
the pair
a3#(a1(a),a,x) f#(x,a) (7)
could be deleted.

1.1.1.1.1 Narrowing Processor

We consider all narrowings of the pair below position ε to get the following set of pairs
a3#(a1(a),a,a) f#(a,a1(a)) (9)
a3#(a1(a),a,a1(x0)) f#(a1(x0),a2(x0,a)) (10)
a3#(a1(a),a,a2(x0,x1)) f#(a2(x0,x1),a3(x0,x1,a)) (11)

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.