Theorem 3anibar 1168

Description: Remove a hypothesis from the second member of a biconditional. (Contributed by FL, 22-Jul-2008.)

Hypothesis

Ref	Expression
3anibar.1	⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ (𝜒 ∧ 𝜏)))

Assertion

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

Proof of Theorem 3anibar

Step	Hyp	Ref	Expression
1		3anibar.1	. 2 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ (𝜒 ∧ 𝜏)))
2		simp3 1002	. . 3 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜒)
3	2	biantrurd 305	. 2 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜏 ↔ (𝜒 ∧ 𝜏)))
4	1, 3	bitr4d 191	1 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   ∧ w3a 981

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This theorem depends on definitions:  df-bi 117  df-3an 983

This theorem is referenced by:  frecsuclem  6492  shftfibg  11131  neiint  14617