Theorem 3anibar 1325

Description: Remove a hypothesis from the second member of a biimplication. (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 1134	. 2 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜒)
2		3anibar.1	. 2 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ (𝜒 ∧ 𝜏)))
3	1, 2	mpbirand 705	1 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398   ∧ w3a 1083

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 209  df-an 399  df-3an 1085

This theorem is referenced by:  cpmatel  21321  neiint  21714  islinindfiss  44512