# Stalmarck's method: x ≡ x → z, does z have to be true?

I have been researching Ståmarck's method 1. In the paper cited here, some rules are given. Rules are made of triplets (x, y, z) such that:

y $$\to$$ z $$\equiv$$ x

where x, y and z are booleans which values are either 0 (false) or 1 (true). And rules then infer a consequence (bottom part of the rule).

For example:

$$\frac{(x, y, 1)}{x/1}$$

means that given a triplet (x, y, 1), x has to be 1 (true). This makes sense because if the expression is $$y \to 1$$, then regardless of the value of y, the expression will be true (1). Now my question is, given the following rule:

$$\frac{(x, x, z)}{x/1}$$

Why is it not possible to infer that z = 1 as well? If the truth table of this expression is drawn, this is the result (bear in mind that the result has to be the same value as x for it to make sense):

x | z | x -> z ($$\equiv x$$)
0 | 0 | 1
0 | 1 | 1
1 | 0 | 0
1 | 1 | 1

Clearly, the only line that does not immediately lead to a contradiction is the last one, and in this case z = 1 as well.

Yes, you are right. Having $$x \leftrightarrow (x \rightarrow z)$$ one can deduce that $$x\equiv 1$$ and $$z \equiv 1$$.
In the paper you cite (the link is to another one btw.), rule 6 allows only to deduce that $$x\equiv 1$$ holds.
But then you have the triplet $$(1,1,z)$$ and you can apply rule 4 $$\frac{(x,1,z)}{x/z}$$ to deduce $$z\equiv x$$, i.e., $$z \equiv 1$$ holds, too.
I have also seen other publications that include the conclusion $$z \equiv 1$$ directly in rule 6.