Theorem imdistan 448

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		pm5.42 320	. 2 ⊢ ((𝜑 → (𝜓 → 𝜒)) ↔ (𝜑 → (𝜓 → (𝜑 ∧ 𝜒))))
2		impexp 263	. 2 ⊢ (((𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜒)) ↔ (𝜑 → (𝜓 → (𝜑 ∧ 𝜒))))
3	1, 2	bitr4i 187	1 ⊢ ((𝜑 → (𝜓 → 𝜒)) ↔ ((𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜒)))

This proof depends on syntax axioms:   → wi 4   ∧ wa 104   ↔ wb 105

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This proof depends on definitions:  df-bi 117

This theorem is used by:  imdistand  451  pm5.3  479  rmoim  3027  ss2rab  3324  bezoutlembi  12782