Theorem 3anibar 1327

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		simp3 1136	. 2 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜒)
2		3anibar.1	. 2 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ (𝜒 ∧ 𝜏)))
3	1, 2	mpbirand 703	1 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   ∧ w3a 1085

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 206  df-an 396  df-3an 1087

This theorem is referenced by:  cpmatel  21768  neiint  22163  islinindfiss  45679