a. Converse. (Q ==>
P) If I carry
an umbrella, then it is raining.
b. State the contrapositive. (~Q ==> ~P, where ~ means "not") If
I do not carry an umbrella, then it is not raining.
c. The original is equivalent to b. (and both are also equivalent to ~P or
Q). (parts a and b together make P <==> Q;
equivalence of P,Q; P iff Q)
(By the way, part a is equivalent to part a's contrapositve, ~P ==>
~Q, but that a. Prove S with a truth table 
the required expression is always true:
Recall P ==> Q is only false if P is true and Q is false.
A

B

C

A or B

(A or B)==> C

B ==> C

((A or B)==> C) ==> (B
==> C)

F

F

F

F

T

T

T

F

F

T

F

T

T

T

F

T

F

T

F

F

T

F

T

T

T

T

T

T

T

F

F

T

F

T

T

T

F

T

T

T

T

T

T

T

F

T

F

F

T

T

T

T

T

T

T

T

b. Basic argument: To prove P ==> Q, assume P, and
derive Q.
1. To show ((A or B)==> C) ==> (B ==> C), assume (A or B)
==> C
1a. We need to show B ==> C.
2. To show B ==> C, assume B
2a. We need to show C
3. Since B is true by 2, A or B is true.
4. Since A or B is true, C is true by 1, so 2a and 1a are true.
Another approach is using pure Boolean algebra: Here 'not' is written ~
((A or B)==> C) ==> (B ==> C)
= ~((A or B)==> C) or (B ==> C) substituting for P==>Q, ~P or Q
= ~(~(A or B)or C) or (~B or C) twice more
= ((A or B) and ~C) or ~B or C by Demorgan's law
= (((A or B) and ~C) or C) or ~B regrouping or's
= ((A or B or C) and (~C or C)) or ~B distributing C
= ((A or B or C) and true) or ~B
= (A or B or C) or ~B
= A or C or (B or ~B)
= A or C or true = true