Given:  NONE
Infer:  P v ~P [Law of the Excluded Middle]

1.	Assume ~(P v ~P)
	________________
2.		Assume P
		_________
3.		P (Assumption 2)
4.		P v ~P (3)
5.		~(P v ~P) (Assumption 1)
6.	~P (Contradiction (2-(4,5))
7.	P v ~P (6)
8.	~(P v ~P) (Assumption 1)
9. ~~(P v ~P) (Contradiction (1-(7,8))
10. P v ~P (9)



Given:
P v Q
~P v Q
Infer:
Q

1. Given P v Q
2. Given ~P v Q
   _____________
3. 	Assume Q
	_________
4.	Q (Assumption)

5.	Assume P
        ________
6.		Assume Q
                ________
7.		Q (Assumption)

8.		Assume ~P
		_________
9.			Assume ~Q
			_________
10.			P (Assumption 5)
11.			~P (Assumption 8)
12.		~~Q (Contradiction, 9-(10,11))
13.		Q (12)
14.	~P v Q (Given 2)
15.	Q (14,8-13,6-7)
16.  P v Q (Given 1)
17.  Q (16,5-15,3-4)










If I know P v Q, AND I can assuming P deduce R, AND I can assuming Q deduce R, then I can conclude R.


==>:  "implies".  Sometimes it's written as a sideways horseshoe.  But generally, it's some sort of shape pointing from the left to the right.z

You might say P implies Q, or if P then Q.  Basically, it says that whenever P is true, Q also is.

Rules for ==>:

7.  If you know P==>Q  AND you know P, then you also know Q.  (Modus ponens).

8.  If, under the assumption of P, I can derive Q, then I can conclude P==>Q.

Given:
P ==> Q
~Q
Infer:
~P   [Modus Tollens]

1.  Given P==>Q
2.  Given ~Q
    ___________
3.	Assume P
	________
4.	P (Assumption 3)
5.	P==>Q (Given 1)
6.	Q (4,5)
7.	~Q (Given 2)
8.  ~P (Contradiction (3-(6,7))



Given:
P==>Q
Infer:
~P v Q

1.  Given P==>Q
    ___________
2.	Assume P
	________
3.	P==>Q (Given 1)
4.	P (Assumption 2)
5.	Q (3,4)
6.	~P v Q (5)

7.	Assume ~P
        _________
8.	~P (Assumption 7)
9.	~P v Q (8)

10.	Assume ~(P v ~P)
	________________
11.		Assume P
		_________
12.		P (Assumption 11)
13.		P v ~P (12)
14.		~(P v ~P) (Assumption 10)
15.	~P (Contradiction (11-(13,14))
16.	P v ~P (15)
17.	~(P v ~P) (Assumption 10)
18. ~~(P v ~P) (Contradiction (10-(16,17))
19. P v ~P (18)
20. ~P v Q (18,2-6,7-9)

TRUE = 1
FALSE = 0
P & Q = PQ
P v Q = P+Q-PQ
~P = 1-P
P^2 = P  (since P is 0 or 1).

(P v Q) && (~P v Q)

(P+Q-PQ)((1-P)+Q-(1-P)Q)
(P+Q-PQ)(1-P+Q-Q+PQ)
(P+Q-PQ)(1-P+PQ)
P -P^2 + P^2Q +Q -PQ +PQ^2 -PQ + P^2Q - P^2Q^2
P - P  + PQ + Q -PQ + PQ - PQ + PQ - PQ
Q


<==>:  Biconditional (equivalence)  P <==> Q means that P and Q have the same truth value as each other.  Sometimes we say P if and only Q,
P iff Q.

9.  If I know P <==> Q and I know P, then I can conclude Q.  Likewise, if I know P <==> Q and I know Q, then I can conclude P.

10.  If under the assumption of P, I can derive Q, AND, under the
     assumption of Q, I can derive P, then I can conclude P <==> Q.


Common mistake:
Prove that statement A is true iff statement B is true.

So, first I assume A, and then I derive B.  PIECE OF CAKE.

So, then I assume ~B and then derive ~A.  ALSO REALLY EASY.

NO NO NO NO NO NO NO