Episode 04 – Formal equivalences

De Morgan's laws. Theorem. ¬(p ∧ q) ⇐⇒ (¬p ∨ ¬q). Proof. Let's look at the truth table of each proposition. p q p ∧ q. ¬(p ∧ q) ¬p. ¬q. ¬p ∨ ¬q. 0. 0. 0. 1. 1. 1. 1.
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Episode 04 – Formal equivalences European section – Season 2

Episode 04 – Formal equivalences

De Morgan’s laws Theorem ¬(p ∧ q) ⇐⇒ (¬p ∨ ¬q)

Episode 04 – Formal equivalences

De Morgan’s laws Theorem ¬(p ∧ q) ⇐⇒ (¬p ∨ ¬q) Proof. Let’s look at the truth table of each proposition. p 0 0 1 1

q 0 1 0 1

p∧q 0 0 0 1

¬(p ∧ q) 1 1 1 0

¬p 1 1 0 0

¬q 1 0 1 0

¬p ∨ ¬q 1 1 1 0

Episode 04 – Formal equivalences

De Morgan’s laws Theorem ¬(p ∧ q) ⇐⇒ (¬p ∨ ¬q) Proof. Let’s look at the truth table of each proposition. p 0 0 1 1

q 0 1 0 1

p∧q 0 0 0 1

¬(p ∧ q) 1 1 1 0

¬p 1 1 0 0

¬q 1 0 1 0

¬p ∨ ¬q 1 1 1 0

As ¬(p ∧ q) and ¬p ∨ ¬q have the same truth table, they are formally equivalent.

Episode 04 – Formal equivalences

De Morgan’s laws Theorem ¬(p ∨ q) ⇐⇒ (¬p ∧ ¬q)

Episode 04 – Formal equivalences

De Morgan’s laws Theorem ¬(p ∨ q) ⇐⇒ (¬p ∧ ¬q) Proof. Let’s look at the truth table of each proposition. p 0 0 1 1

q 0 1 0 1

p∨q 0 1 1 1

¬(p ∨ q) 1 0 0 0

¬p 1 1 0 0

¬q 1 0 1 0

¬p ∧ ¬q 1 0 0 0

Episode 04 – Formal equivalences

De Morgan’s laws Theorem ¬(p ∨ q) ⇐⇒ (¬p ∧ ¬q) Proof. Let’s look at the truth table of each proposition. p 0 0 1 1

q 0 1 0 1

p∨q 0 1 1 1

¬(p ∨ q) 1 0 0 0

¬p 1 1 0 0

¬q 1 0 1 0

¬p ∧ ¬q 1 0 0 0

As ¬(p ∨ q) and ¬p ∧ ¬q have the same truth table, they are formally equivalent.

Episode 04 – Formal equivalences

Conditional Theorem (p → q) ⇐⇒ (¬q → ¬p) ⇐⇒ (¬p ∨ q)

Episode 04 – Formal equivalences

Conditional Theorem (p → q) ⇐⇒ (¬q → ¬p) ⇐⇒ (¬p ∨ q) Proof. The first formal equivalence has been proven in the previous session, so we just have to look at the truth tables of p → q and ¬p ∨ q. p 0 0 1 1

q 0 1 0 1

p→q 1 1 0 1

¬p 1 1 0 0

¬p ∨ q 1 1 0 1

Episode 04 – Formal equivalences

Conditional Theorem (p → q) ⇐⇒ (¬q → ¬p) ⇐⇒ (¬p ∨ q) Proof. The first formal equivalence has been proven in the previous session, so we just have to look at the truth tables of p → q and ¬p ∨ q. p 0 0 1 1

q 0 1 0 1

p→q 1 1 0 1

¬p 1 1 0 0

¬p ∨ q 1 1 0 1

As p → q and ¬p ∨ q have the same truth table, they are formally equivalent.

Episode 04 – Formal equivalences

Biconditional Theorem (p ↔ q) ⇐⇒ (¬q ↔ ¬p)

Episode 04 – Formal equivalences

Biconditional Theorem (p ↔ q) ⇐⇒ (¬q ↔ ¬p) Proof. Once again, we just have to look at the truth tables of the two propositions. p 0 0 1 1

q 0 1 0 1

p↔q 1 0 0 1

¬p 1 1 0 0

¬q 1 0 1 0

¬p ↔ ¬q 1 0 0 1

Episode 04 – Formal equivalences

Biconditional Theorem (p ↔ q) ⇐⇒ (¬q ↔ ¬p) Proof. Once again, we just have to look at the truth tables of the two propositions. p 0 0 1 1

q 0 1 0 1

p↔q 1 0 0 1

¬p 1 1 0 0

¬q 1 0 1 0

¬p ↔ ¬q 1 0 0 1

As p ↔ q and ¬p ↔ ¬q have the same truth table, they are formally equivalent.

Episode 04 – Formal equivalences