Theorem 3anibar 1123

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		3anibar.1	. 2 ⊢ ((φ ∧ ψ ∧ χ) → (θ ↔ (χ ∧ τ)))
2		simp3 957	. . 3 ⊢ ((φ ∧ ψ ∧ χ) → χ)
3	2	biantrurd 494	. 2 ⊢ ((φ ∧ ψ ∧ χ) → (τ ↔ (χ ∧ τ)))
4	1, 3	bitr4d 247	1 ⊢ ((φ ∧ ψ ∧ χ) → (θ ↔ τ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358   ∧ w3a 934

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177  df-an 360  df-3an 936

This theorem is used by: (None)