Theorem imdistand 575

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 572	. 2 ⊢ ((𝜓 → (𝜒 → 𝜃)) ↔ ((𝜓 ∧ 𝜒) → (𝜓 ∧ 𝜃)))
3	1, 2	sylib 219	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 208  df-an 397

This theorem is referenced by:  imdistanda  576  a2and  851  reximdvai  3150  unblem1  9192  cfub  10162  lbzbi  12877  ltslpss  27918  cusgredgex  35350  poimirlem32  38019  ispridl2  38405  ispridlc  38437  lnr2i  43561  rfovcnvf1od  44448