Theorem 3impexp 1326

Description: impexp 250 with a 3-conjunct antecedent. (Contributed by Alan Sare, 31-Dec-2011.)

Assertion

Ref	Expression
3impexp	⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ↔ (𝜑 → (𝜓 → (𝜒 → 𝜃))))

Proof of Theorem 3impexp

Step	Hyp	Ref	Expression
1		id 19	. . 3 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃))
2	1	3expd 1121	. 2 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) → (𝜑 → (𝜓 → (𝜒 → 𝜃))))
3		id 19	. . 3 ⊢ ((𝜑 → (𝜓 → (𝜒 → 𝜃))) → (𝜑 → (𝜓 → (𝜒 → 𝜃))))
4	3	3impd 1118	. 2 ⊢ ((𝜑 → (𝜓 → (𝜒 → 𝜃))) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃))
5	2, 4	impbii 117	1 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ↔ (𝜑 → (𝜓 → (𝜒 → 𝜃))))

Syntax hints:   → wi 4   ↔ wb 98   ∧ w3a 885

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101

This theorem depends on definitions:  df-bi 110  df-3an 887

This theorem is referenced by:  3impexpbicom  1327