Theorem imdistan 567

Description: Distribution of implication with conjunction. (Contributed by NM, 31-May-1999.) (Proof shortened by Wolf Lammen, 6-Dec-2012.)

Assertion

Ref	Expression
imdistan	⊢ ((𝜑 → (𝜓 → 𝜒)) ↔ ((𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜒)))

Proof of Theorem imdistan

Step	Hyp	Ref	Expression
1		pm5.42 543	. 2 ⊢ ((𝜑 → (𝜓 → 𝜒)) ↔ (𝜑 → (𝜓 → (𝜑 ∧ 𝜒))))
2		impexp 450	. 2 ⊢ (((𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜒)) ↔ (𝜑 → (𝜓 → (𝜑 ∧ 𝜒))))
3	1, 2	bitr4i 277	1 ⊢ ((𝜑 → (𝜓 → 𝜒)) ↔ ((𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜒)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ 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 206  df-an 396

This theorem is referenced by:  imdistand  570  rmoim  3670  ss2rab  4000  marypha2lem3  9126  ismhp3  21243  inxpss3  36376