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

[physicsuser]

need to prove this

<br /> \frac{\urcorner P \equiv false}{P \equiv true}<br />

here is what I did
using Leibniz

<br /> \frac{X \equiv Y}{E[z:=X] \equiv E[z:=Y]}<br />

<br /> X=\urcorner P<br />

<br /> Y=false<br />

<br /> E:\urcorner z<br />

<br /> z=z<br />

<br /> \frac{\urcorner P \equiv false}{\urcorner\urcorner P \equiv \urcorner false}<br />

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

<br /> \frac{\urcorner P \equiv false}{P \equiv true}<br />

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!