Theorem imdistan 578

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		pm5.42 553	. 2 ⊢ ((𝜑 → (𝜓 → 𝜒)) ↔ (𝜑 → (𝜓 → (𝜑 ∧ 𝜒))))
2		impexp 456	. 2 ⊢ (((𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜒)) ↔ (𝜑 → (𝜓 → (𝜑 ∧ 𝜒))))
3	1, 2	bitr4i 281	1 ⊢ ((𝜑 → (𝜓 → 𝜒)) ↔ ((𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜒)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ 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:  imdistand  581  rmoim  3705  ss2rab  4024  ss2rabd  4027  marypha2lem3  9404  ismhp3  22357  inxpss3  39029