Claim: (\forall m, n \in \mathbb{Z})(2m \neq 2n+1)
Proof: AFSOC (\exists m,n \in \mathbb{Z})(2m = 2n+1), then m-n=\frac{1}{2}. However, m-n \in \mathbb{Z}. There is a contradiction. \Rightarrow\!\Leftarrow

Existential Statement

Claim: (\exists x \in S)P(x)
Direct Proof: proof by demonstration that some number cannot be in integer

Disprove by Counter Example for Union 1

Claim: A\nsubseteq B
Proof: prove by showing (\exists a \in A) s.t. a\notin B.
Proof:

Claim: There exist 2 irrational numbers a and b such that a^b \in \mathbb{Q}
Proof: let a=b=\sqrt{2}

case 1: \sqrt{2}^{\sqrt{2}} \in \mathbb{Q} - satisfy claim.

case 2: \sqrt{2}^{\sqrt{2}} \notin \mathbb{Q}, then \sqrt{2}^{\sqrt{2}^\sqrt{2}}=\sqrt{2}^2 \in \mathbb{Q} where \sqrt{2}^{\sqrt{2}} \notin \mathbb{Q} - satisfy claim.

Indirect Proof

Using \lnot (\exists x \in S)P(x) \equiv (\forall x \in S)\lnot P(x)
Claim: Let (n\in \mathbb{Z}^{+}) \land (y_1,...,y_n \in \mathbb{R}), (\exists i \in [n])(y_i \geq \sum_{j=1}^n \frac{y_j}{n})
Proof: AFSOC (\forall i \in [n])(y_i < \sum_{j=1}^n \frac{y_j}{n}), then summing up all y_i, \sum_{i=1}^n y_i < n\sum_{j=1}^n \frac{y_i}{n} = \sum_{i=j}^n y_j, which contradict with \sum_{i=1}^n y_i = \sum_{i=j}^n y_j. \therefore (\exists i \in [n])(y_i \geq \sum_{j=1}^n \frac{y_j}{n})

Contradiction Claims

Claim: P \implies Q
Proof:

Direct Proof: assume P, show Q holds

Contrapositive: show \lnot Q \implies \lnot P

Contradiction: show \lnot(P\implies Q) \equiv (P\land \lnot Q) by AFSOC P \land \lnot Q.

Even Odd

Claim: (\forall n \in \mathbb{Z})((\exists m\in \mathbb{Z})(n=2m)\implies (\exists l \in \mathbb{Z})(n^2 = 2l))
Proof: let n\in \mathbb{Z} be even, s.t. n=2m for some m\in \mathbb{Z}. Fix m. Then x^2 = (2m)^2 = 4m^2 = 2(2m^2), and since (2m^2 \in \mathbb{Z}), n^2 is even.