Theorem anandir 690

Description: Distribution of conjunction over conjunction. (Contributed by NM, 24-Aug-1995.)

Assertion

Ref	Expression
anandir	⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ((𝜑 ∧ 𝜒) ∧ (𝜓 ∧ 𝜒)))

Proof of Theorem anandir

Step	Hyp	Ref	Expression
1		anidm 575	. . 3 ⊢ ((𝜒 ∧ 𝜒) ↔ 𝜒)
2	1	anbi2i 635	. 2 ⊢ (((𝜑 ∧ 𝜓) ∧ (𝜒 ∧ 𝜒)) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
3		an4 669	. 2 ⊢ (((𝜑 ∧ 𝜓) ∧ (𝜒 ∧ 𝜒)) ↔ ((𝜑 ∧ 𝜒) ∧ (𝜓 ∧ 𝜒)))
4	2, 3	bitr3i 280	1 ⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ((𝜑 ∧ 𝜒) ∧ (𝜓 ∧ 𝜒)))

This proof depends on syntax axioms:   ↔ 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:  anandi3r  1120  disjxun  5100  fununi  6616  imadif  6625  elfzuzb  13654  frgr3v  30882  5oalem3  32264  5oalem5  32266  refrelredund4  39745  nzin  45396  un2122  45866