Certification Problem

Input (TPDB TRS_Standard/Zantema_05/jw30)

The rewrite relation of the following TRS is considered.

f(f(a,a),x) → f(f(x,a),f(a,f(a,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)


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

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.
a2#(a,x) → f#(f(x,a),a1(a1(a))) (5)
a2#(a,x) → f#(x,a) (6)
f#(a1(x0),y1) → a2#(x0,y1) (7)

1.1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[a2#(x1, x2)] =
1
0
+
1 0
0 0
· x1 +
0 1
0 0
· x2
[a] =
1
0
[f#(x1, x2)] =
1
0
+
0 1
0 0
· x1 +
0 1
0 0
· x2
[f(x1, x2)] =
0
0
+
0 0
0 0
· x1 +
0 0
1 0
· x2
[a1(x1)] =
0
0
+
0 0
1 0
· x1
[a2(x1, x2)] =
0
0
+
0 0
0 0
· x1 +
0 0
1 0
· x2
the pair
a2#(a,x) → f#(x,a) (6)
could be deleted.

1.1.1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[a2#(x1, x2)] =
0
0
+
0 0
0 0
· x1 +
1 0
0 0
· x2
[a] =
1
0
[f#(x1, x2)] =
0
0
+
0 1
0 0
· x1 +
1 0
0 0
· x2
[f(x1, x2)] =
0
0
+
0 0
1 0
· x1 +
0 0
0 0
· x2
[a1(x1)] =
0
1
+
0 0
0 0
· x1
[a2(x1, x2)] =
0
0
+
0 0
0 0
· x1 +
0 0
0 0
· x2
the pair
f#(a1(x0),y1) → a2#(x0,y1) (7)
could be deleted.

1.1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals
[f(x1, x2)] = 1 · x1 + 2 · x2
[a] = 0
[a1(x1)] = 2 · x1
[a2(x1, x2)] = 2 · x1 + 2 · x2
[a2#(x1, x2)] = 1 · x1 + 2 · x2
[f#(x1, x2)] = 2 · x1 + 2 · x2
together with the usable rules
f(a,y1) → a1(y1) (2)
f(a1(x0),y1) → a2(x0,y1) (3)
a2(a,x) → f(f(x,a),a1(a1(a))) (4)
(w.r.t. the implicit argument filter of the reduction pair), the pair
a2#(a,x) → f#(f(x,a),a1(a1(a))) (5)
and no rules could be deleted.

1.1.1.1.1.1.1 P is empty

There are no pairs anymore.