Theorem imdistand 673

Description: Distribution of implication with conjunction (deduction rule). (Contributed by NM, 27-Aug-2004.)

Hypothesis

Ref	Expression
imdistand.1	⊢ (φ → (ψ → (χ → θ)))

Assertion

Ref	Expression
imdistand	⊢ (φ → ((ψ ∧ χ) → (ψ ∧ θ)))

Proof of Theorem imdistand

Step	Hyp	Ref	Expression
1		imdistand.1	. 2 ⊢ (φ → (ψ → (χ → θ)))
2		imdistan 670	. 2 ⊢ ((ψ → (χ → θ)) ↔ ((ψ ∧ χ) → (ψ ∧ θ)))
3	1, 2	sylib 188	1 ⊢ (φ → ((ψ ∧ χ) → (ψ ∧ θ)))

This proof depends on syntax axioms:   → wi 4   ∧ wa 358

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

This theorem is used by:  imdistanda  674  fconstfv  5457  nchoicelem19  6308