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

[physicsuser]

need to prove this

[tex]
\frac{\urcorner P \equiv false}{P \equiv true}
[/tex]

here is what I did
using Leibniz

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

[tex]
X=\urcorner P
[/tex]

[tex]
Y=false
[/tex]

[tex]
E:\urcorner z
[/tex]

[tex]
z=z
[/tex]

[tex]
\frac{\urcorner P \equiv false}{\urcorner\urcorner P \equiv \urcorner false}
[/tex]

since [tex]\urcorner\urcorner P \equiv P[/tex]
and [tex]\urcorner false \equiv true [/tex]

[tex]
\frac{\urcorner P \equiv false}{P \equiv true}
[/tex]

is this a proof?

 

[anathema]

Yes, this is a valid proof using Leibniz's rule. You have correctly substituted \urcorner P for X and false for Y in the formula, and then applied the rule to show that \urcorner P \equiv false implies P \equiv true. Well done!

 

[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!