# Is true * true = true in Separation Logic?

I am trying to show that the following interference is unsound in terms of Separation Logic:

$$(p_0 \implies p_1) \implies ((p_0 * q) \implies (p_1 * q))$$

I came up with the following values for a counter example: $$p_0 = true$$, $$p_1 = x \mapsto 1$$, $$q = true$$. The heap contains just one mapping of $$x \mapsto 1$$.

The idea is that $$(p_1 * q)$$ cannot be true because $$p_1$$ already describes the whole heap and there is nothing left for $$q$$.

I am not sure, however, why $$true * true$$ would hold if there is only one element in the heap. Shouldn’t it be false as those two $$true$$s cannot claim that single heap element at the same time?