\( \forall P, Q: \; (P \Rightarrow Q) \; \equiv \; ((P \wedge Q) \equiv P) \)
The English version of the question, in case you are not familiar with
the notation:
for all \(P, Q\):
\(P\) IMPLIES \(Q\) is equivalent to
\(P\) AND \(Q\) is equivalent to \(P\)
Here are three different ways of answering they quesion.
Case: \(P, Q = True, True\):
\(
P \Rightarrow Q \quad \equiv \quad (P \wedge Q \; \equiv \; P)
\)
\(
T \Rightarrow T \quad \equiv \quad (T \wedge T \; \equiv \; T)
\)
\(
T \quad \equiv \quad (T \; \equiv \; T)
\)
\(
T \quad \equiv \quad T
\)
\(
T
\)
Case: \(P, Q = True, False\):
\(
P \Rightarrow Q \quad \equiv \quad (P \wedge Q \; \equiv \; P)
\)
\(
T \Rightarrow F \quad \equiv \quad (T \wedge F \; \equiv \; T)
\)
\(
F \quad \equiv \quad (F \; \equiv \; T)
\)
\(
F \quad \equiv \quad F
\)
\(
T
\)
Case: \(P, Q = False, True\):
\(
P \Rightarrow Q \quad \equiv \quad (P \wedge Q \; \equiv \; P)
\)
\(
F \Rightarrow Q \quad \equiv \quad (F \wedge T \; \equiv \; F)
\)
\(
T \quad \equiv \quad (F \; \equiv \; F)
\)
\(
T \quad \equiv \quad T
\)
\(
T
\)
Case: \(P, Q = False, False\):
\(
P \Rightarrow Q \quad \equiv \quad (P \wedge Q \; \equiv \; P)
\)
\(
F \Rightarrow F \quad \equiv \quad (F \wedge F \; \equiv \; F)
\)
\(
T \quad \equiv \quad (F \; \equiv \; F)
\)
\(
T \quad \equiv \quad T
\)
\(
T
\)
K. Mani Chandy, Emeritus Simon Ramo Professor, California Institute of Technology