Certification Problem

Input (TPDB TRS_Standard/Zantema_05/z05)

The rewrite relation of the following TRS is considered.

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

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)
c is mapped to c, c1(x1)
b is mapped to b, b1(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(a1(x)) c1(b1(x)) (7)
b1(b1(x)) a1(c1(x)) (8)
c1(c1(x)) b1(a1(x)) (9)
f(a,y1) a1(y1) (4)
f(c,y1) c1(y1) (5)
f(b,y1) b1(y1) (6)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a1(x1)] = 2 + 2 · x1
[c1(x1)] = 2 + 2 · x1
[b1(x1)] = 2 + 2 · x1
[f(x1, x2)] = 2 + 1 · x1 + 2 · x2
[a] = 0
[c] = 1
[b] = 2
all of the following rules can be deleted.
f(c,y1) c1(y1) (5)
f(b,y1) b1(y1) (6)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a1(x1)] = 2 + 2 · x1
[c1(x1)] = 2 + 2 · x1
[b1(x1)] = 2 + 2 · x1
[f(x1, x2)] = 1 + 2 · x1 + 2 · x2
[a] = 2
all of the following rules can be deleted.
f(a,y1) a1(y1) (4)

1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a1#(a1(x)) c1#(b1(x)) (10)
a1#(a1(x)) b1#(x) (11)
b1#(b1(x)) a1#(c1(x)) (12)
b1#(b1(x)) c1#(x) (13)
c1#(c1(x)) b1#(a1(x)) (14)
c1#(c1(x)) a1#(x) (15)

1.1.1.1.1 Monotonic Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a1(x1)] = 1 + 2 · x1
[c1(x1)] = 1 + 2 · x1
[b1(x1)] = 1 + 2 · x1
[a1#(x1)] = 1 · x1
[c1#(x1)] = 1 · x1
[b1#(x1)] = 1 · x1
the pairs
a1#(a1(x)) b1#(x) (11)
b1#(b1(x)) c1#(x) (13)
c1#(c1(x)) a1#(x) (15)
and no rules could be deleted.

1.1.1.1.1.1 Narrowing Processor

We consider all narrowings of the pair below position 1 to get the following set of pairs
a1#(a1(b1(x0))) c1#(a1(c1(x0))) (16)

1.1.1.1.1.1.1 Narrowing Processor

We consider all narrowings of the pair below position 1 to get the following set of pairs
b1#(b1(c1(x0))) a1#(b1(a1(x0))) (17)

1.1.1.1.1.1.1.1 Narrowing Processor

We consider all narrowings of the pair below position 1 to get the following set of pairs
c1#(c1(a1(x0))) b1#(c1(b1(x0))) (18)

1.1.1.1.1.1.1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
[a1#(x1)] =
-∞
-∞
-∞
+
0 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a1(x1)] =
-∞
-∞
1
+
0 -∞ 0
0 -∞ 0
1 1 0
· x1
[b1(x1)] =
0
1
0
+
0 0 -∞
1 0 0
-∞ 1 0
· x1
[c1#(x1)] =
-∞
-∞
-∞
+
0 0 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[c1(x1)] =
-∞
1
-∞
+
0 0 0
1 -∞ 0
1 0 0
· x1
[b1#(x1)] =
-∞
-∞
-∞
+
0 0 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
the pair
a1#(a1(b1(x0))) c1#(a1(c1(x0))) (16)
could be deleted.

1.1.1.1.1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.