p --> q and q --> p are not equivalent p --> q and q --> (not)p are equivalent The truth table shows this. pq p --> q q -->(not)p f f t t f t t t t f f f t t t t
p > q~qTherefore, ~p| p | q | p > q | ~q | ~p || t | t | t | f | f || t | f | t | t | f || f | t | t | f | t || f | f | t | t | t |
4t - 2p
P Q (/P or /Q) T T F T F T F T T F F T
Proof: P{T>n+m/T>n}=P{T>n+m,T>n}/P{T>n} (Bayes theorem) =P{T>n+m}/P{T>n} =((1-p)^(n+m))/(1-p)^n = (1-p)^(n+m-n) = (1-p)^m (1-p)^m = {T>m} So T>m has the same probability as T>m+n given that T>n, which means it doesn't care (or don't remember) that n phases had passed.
A. P. T. James was born in 1908.
T. P. Gill died in 1931.
T. P. O'Connor died in 1929.
T. P. O'Connor was born in 1848.
T. P. Rajalakshmi died in 1964.
T. P. Rajalakshmi was born in 1911.
P. T. Selbit died in 1938.