Why the material conditional has the truth table it does
(assuming that it has a truth table!):
P
|
–>
|
Q
|
T
|
T
|
T
|
T
|
F
|
F
|
F
|
T
|
T
|
F
|
T
|
F
|
(i) It's a logical truth that if both A and B are true,
then A is true; so
(ii) (A ^ B) –> A is a tautology – T in every row of its
truth table.
A
|
B
|
|
|
(A ^ B)
|
–>
|
A
|
T
|
T
|
|
|
T
|
T
|
T
|
T
|
F
|
|
|
F
|
T
|
T
|
F
|
T
|
|
|
F
|
T
|
F
|
F
|
F
|
|
|
F
|
T
|
F
|
(iii) From the first row we see T –> T is T,
(iv) from the second row we see F –> T is T, and
(v) from the third row we see F –> F is T.
(iv) That only leaves the row for T –> F; but since P –>
Q is not a tautology, that row must be F.
QED
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