Proving Leibniz Logic: \frac{\urcorner P \equiv false}{P \equiv true}

[physicsuser]

need to prove this

$$\frac{\urcorner P \equiv false}{P \equiv true}$$

here is what I did
using Leibniz

$$\frac{X \equiv Y}{E[z:=X] \equiv E[z:=Y]}$$

$$X=\urcorner P$$

$$Y=false$$

$$E:\urcorner z$$

$$z=z$$

$$\frac{\urcorner P \equiv false}{\urcorner\urcorner P \equiv \urcorner false}$$

since $$\urcorner\urcorner P \equiv P$$
and $$\urcorner false \equiv true$$

$$\frac{\urcorner P \equiv false}{P \equiv true}$$

is this a proof?

 

[wais]

Yes, this is a valid proof using Leibniz logic. You have correctly applied the substitution principle and used the definitions of negation and false to show that \urcorner P \equiv false implies P \equiv true. Well done!