Theorem imdistand 571

Description: Distribution of implication with conjunction (deduction form). (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 568	. 2 ⊢ ((𝜓 → (𝜒 → 𝜃)) ↔ ((𝜓 ∧ 𝜒) → (𝜓 ∧ 𝜃)))
3	1, 2	sylib 217	1 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) → (𝜓 ∧ 𝜃)))

Syntax hints:   → wi 4   ∧ wa 396

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 397

This theorem is referenced by:  imdistanda  572  a2and  842  reximdvai  3200  unblem1  9066  cfub  10005  lbzbi  12676  cusgredgex  33083  sltlpss  34087  poimirlem32  35809  ispridl2  36196  ispridlc  36228  lnr2i  40941  rfovcnvf1od  41612