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At begin I would like to say that this question seems to be long and complex - no, here is only one short and straightforward question. This is so long, because I did my best to exactly reflect my doubts.

Lets consider formula: $$(p\to q)\to (\neg p \vee q)$$ Prove that it is not tautology in intutionistical logic

On the whole it isn't hard for me. I have only one question:
I think that in order to show non-tautology it is sufficient to show any model such that in this model there isn't forced considered formula.

For given example: If I can consider chosen (not arbitrary/every) model to show non-tautology ? For example, I begin with something like that:
$$w_1\vdash p $$ $$w_1\not\vdash q $$ $$w_2\vdash q $$ $$w_2\not\vdash p$$
It is chosen by me model, I can continue to prove that given formula is not tautology. Is it end of proof? I afraid of that I choose this model, maybe I should consider all models ?

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    You only need to define one Kripke model in which the formula fails to hold to show that the formula is not intuitionistically valid. But to define the model you need to give the accessibility relation between the worlds as well as the forcing relation. – 2017-02-21
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    So one model sufficies to show non-axiomatization ? – 2017-02-21
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    One model that does not satisfy your formula is sufficient to show that the formula is not intuitionistically valid (i.e., not a "tautology in intuitionistic logic" to use your words). – 2017-02-21

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