Theorem bj-dfbi6 37445

Description: Alternate definition of the biconditional. (Contributed by BJ, 4-Oct-2019.)

Assertion

Ref	Expression
bj-dfbi6	⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ∧ 𝜓)))

Proof of Theorem bj-dfbi6

Step	Hyp	Ref	Expression
1		bj-dfbi5 37444	. 2 ⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 ∨ 𝜓) → (𝜑 ∧ 𝜓)))
2		id 23	. . . 4 ⊢ (((𝜑 ∨ 𝜓) → (𝜑 ∧ 𝜓)) → ((𝜑 ∨ 𝜓) → (𝜑 ∧ 𝜓)))
3		animorr 994	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → (𝜑 ∨ 𝜓))
4	2, 3	impbid1 228	. . 3 ⊢ (((𝜑 ∨ 𝜓) → (𝜑 ∧ 𝜓)) → ((𝜑 ∨ 𝜓) ↔ (𝜑 ∧ 𝜓)))
5		biimp 218	. . 3 ⊢ (((𝜑 ∨ 𝜓) ↔ (𝜑 ∧ 𝜓)) → ((𝜑 ∨ 𝜓) → (𝜑 ∧ 𝜓)))
6	4, 5	impbii 212	. 2 ⊢ (((𝜑 ∨ 𝜓) → (𝜑 ∧ 𝜓)) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ∧ 𝜓)))
7	1, 6	bitri 278	1 ⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ∧ 𝜓)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862

This theorem is used by: (None)