Theorem imdistand 570

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 567	. 2 ⊢ ((𝜓 → (𝜒 → 𝜃)) ↔ ((𝜓 ∧ 𝜒) → (𝜓 ∧ 𝜃)))
3	1, 2	sylib 218	1 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) → (𝜓 ∧ 𝜃)))

Syntax hints:   → wi 4   ∧ wa 395

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 207  df-an 396

This theorem is referenced by:  imdistanda  571  a2and  845  reximdvai  3163  unblem1  9326  cfub  10287  lbzbi  12976  sltlpss  27960  cusgredgex  35106  poimirlem32  37639  ispridl2  38025  ispridlc  38057  lnr2i  43105  rfovcnvf1od  43994