Theorem anandir 689

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 574	. . 3 ⊢ ((𝜒 ∧ 𝜒) ↔ 𝜒)
2	1	anbi2i 634	. 2 ⊢ (((𝜑 ∧ 𝜓) ∧ (𝜒 ∧ 𝜒)) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
3		an4 668	. 2 ⊢ (((𝜑 ∧ 𝜓) ∧ (𝜒 ∧ 𝜒)) ↔ ((𝜑 ∧ 𝜒) ∧ (𝜓 ∧ 𝜒)))
4	2, 3	bitr3i 280	1 ⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ((𝜑 ∧ 𝜒) ∧ (𝜓 ∧ 𝜒)))

Syntax hints:   ↔ wb 209   ∧ wa 400

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 210  df-an 401

This theorem is referenced by:  anandi3r  1118  disjxun  5103  fununi  6600  imadif  6609  elfzuzb  13537  frgr3v  30535  5oalem3  31917  5oalem5  31919  refrelredund4  39230  nzin  44892  un2122  45363