Theorem 3impexpbicom 1327

Description: 3impexp 1326 with biconditional consequent of antecedent that is commuted in consequent. (Contributed by Alan Sare, 31-Dec-2011.)

Assertion

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

Proof of Theorem 3impexpbicom

Step	Hyp	Ref	Expression
1		bicom 128	. . . 4 ⊢ ((𝜃 ↔ 𝜏) ↔ (𝜏 ↔ 𝜃))
2		imbi2 226	. . . . 5 ⊢ (((𝜃 ↔ 𝜏) ↔ (𝜏 ↔ 𝜃)) → (((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏)) ↔ ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜏 ↔ 𝜃))))
3	2	biimpcd 148	. . . 4 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏)) → (((𝜃 ↔ 𝜏) ↔ (𝜏 ↔ 𝜃)) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜏 ↔ 𝜃))))
4	1, 3	mpi 15	. . 3 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏)) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜏 ↔ 𝜃)))
5	4	3expd 1121	. 2 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏)) → (𝜑 → (𝜓 → (𝜒 → (𝜏 ↔ 𝜃)))))
6		3impexp 1326	. . . 4 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜏 ↔ 𝜃)) ↔ (𝜑 → (𝜓 → (𝜒 → (𝜏 ↔ 𝜃)))))
7	6	biimpri 124	. . 3 ⊢ ((𝜑 → (𝜓 → (𝜒 → (𝜏 ↔ 𝜃)))) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜏 ↔ 𝜃)))
8	7, 1	syl6ibr 151	. 2 ⊢ ((𝜑 → (𝜓 → (𝜒 → (𝜏 ↔ 𝜃)))) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏)))
9	5, 8	impbii 117	1 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏)) ↔ (𝜑 → (𝜓 → (𝜒 → (𝜏 ↔ 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: (None)