dekoi
Nov21-04, 11:21 AM
\vdash \left( (\vdash p)\rightarrow(p \pmod 2 \equiv 0) \right)
... means :
\vdash \left( (\vdash p)\rightarrow 1-no \rightarrow yes
Therefore, left side = yes.
Now the right side, (p \pmod 2 \equiv 0) \right) is valid, since 'mod2' = 0 (mod'even'=0, mod'odd'=1 right?).
Am i correct with this information?
(I am sorry for sort of double-posting this thread, but i guess the one in the philosophy forum was out of context).
... means :
\vdash \left( (\vdash p)\rightarrow 1-no \rightarrow yes
Therefore, left side = yes.
Now the right side, (p \pmod 2 \equiv 0) \right) is valid, since 'mod2' = 0 (mod'even'=0, mod'odd'=1 right?).
Am i correct with this information?
(I am sorry for sort of double-posting this thread, but i guess the one in the philosophy forum was out of context).