Theorem 3impexp 1425

Description: impexp 261 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 1214	. 2 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) → (𝜑 → (𝜓 → (𝜒 → 𝜃))))
3		id 19	. . 3 ⊢ ((𝜑 → (𝜓 → (𝜒 → 𝜃))) → (𝜑 → (𝜓 → (𝜒 → 𝜃))))
4	3	3impd 1211	. 2 ⊢ ((𝜑 → (𝜓 → (𝜒 → 𝜃))) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃))
5	2, 4	impbii 125	1 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ↔ (𝜑 → (𝜓 → (𝜒 → 𝜃))))

Syntax hints:   → wi 4   ↔ wb 104   ∧ w3a 968

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116  df-3an 970

This theorem is referenced by:  3impexpbicom  1426