Theorem expcomdg 40841

Description: Biconditional form of expcomd 419. (Contributed by Alan Sare, 22-Jul-2012.) (New usage is discouraged.)

Assertion

Ref	Expression
expcomdg	⊢ ((𝜑 → ((𝜓 ∧ 𝜒) → 𝜃)) ↔ (𝜑 → (𝜒 → (𝜓 → 𝜃))))

Proof of Theorem expcomdg

Step	Hyp	Ref	Expression
1		ancomst 467	. . 3 ⊢ (((𝜓 ∧ 𝜒) → 𝜃) ↔ ((𝜒 ∧ 𝜓) → 𝜃))
2		impexp 453	. . 3 ⊢ (((𝜒 ∧ 𝜓) → 𝜃) ↔ (𝜒 → (𝜓 → 𝜃)))
3	1, 2	bitri 277	. 2 ⊢ (((𝜓 ∧ 𝜒) → 𝜃) ↔ (𝜒 → (𝜓 → 𝜃)))
4	3	imbi2i 338	1 ⊢ ((𝜑 → ((𝜓 ∧ 𝜒) → 𝜃)) ↔ (𝜑 → (𝜒 → (𝜓 → 𝜃))))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 209  df-an 399

This theorem is referenced by: (None)