Quasi Reductive Proof

by ttt2

Input

The rewrite relation of the following conditional TRS is considered.

id(s(x)) id(id(z)) | id(x) ≈ y, id(y) ≈ z
id(0) 0

Proof

1 Unraveling

To prove that the CTRS is quasi-reductive, we show termination of the following unraveled system.

For id(s(x))id(id(z))id(x)yid(y)z we get
id(s(x)) U(id(x),x)
U(y,x) W(id(y))
W(z) id(id(z))
For id(0)0 we get
id(0) 0

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the rationals with delta = 1/64
[0] =
0 0 0
0 0 0
0 0 0
[W(x1)] =
1 1 0
0 1 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[U(x1, x2)] =
1 1 0
0 0 0
1 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
1 0 0
[id(x1)] =
1 1 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 1
1 1 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
the rules
U(y,x) W(id(y))
W(z) id(id(z))
id(0) 0
remain.

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the rationals with delta = 1/64
[0] =
0 0 0
0 0 0
0 0 0
[W(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[U(x1, x2)] =
1 0 1
0 1 1
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
1 0 0
1 0 0
1 0 0
[id(x1)] =
1 0 0
0 0 1
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
the rules
W(z) id(id(z))
id(0) 0
remain.

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the rationals with delta = 1/64
[0] =
1 0 0
1 0 0
1 0 0
[W(x1)] =
1 1 1
1 1 0
0 0 1
· x1 +
1 0 0
1 0 0
1 0 0
[id(x1)] =
1 1 0
0 0 1
0 1 0
· x1 +
0 0 0
1 0 0
0 0 0
the rule
W(z) id(id(z))
remains.

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[W(x1)] = 15 · x1 + -∞
[id(x1)] = 0 · x1 + -∞
all rules could be removed.

1.1.1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.

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