Theorem imdistand 581

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 578	. 2 ⊢ ((𝜓 → (𝜒 → 𝜃)) ↔ ((𝜓 ∧ 𝜒) → (𝜓 ∧ 𝜃)))
3	1, 2	sylib 221	1 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) → (𝜓 ∧ 𝜃)))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401

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

This proof depends on definitions:  df-bi 210  df-an 402

This theorem is used by:  imdistanda  582  a2and  859  reximdvai  3178  unblem1  9259  cfub  10247  lbzbi  12976  ltslpss  28152  cusgredgex  35664  poimirlem32  38360  ispridl2  38747  ispridlc  38779  lnr2i  43901  rfovcnvf1od  44788