Theorem anandi 801

Description: Distribution of conjunction over conjunction. (Contributed by NM, 14-Aug-1995.)

Assertion

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

Proof of Theorem anandi

Step	Hyp	Ref	Expression
1		anidm 625	. . 3 ⊢ ((φ ∧ φ) ↔ φ)
2	1	anbi1i 676	. 2 ⊢ (((φ ∧ φ) ∧ (ψ ∧ χ)) ↔ (φ ∧ (ψ ∧ χ)))
3		an4 797	. 2 ⊢ (((φ ∧ φ) ∧ (ψ ∧ χ)) ↔ ((φ ∧ ψ) ∧ (φ ∧ χ)))
4	2, 3	bitr3i 242	1 ⊢ ((φ ∧ (ψ ∧ χ)) ↔ ((φ ∧ ψ) ∧ (φ ∧ χ)))

This proof depends on syntax axioms:   ↔ wb 176   ∧ wa 358

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177  df-an 360

This theorem is used by:  inrab  3528  uniin  3912  fin  5247  inpreima  5410  fununiq  5518  ndmovdistr  5620  trtxp  5782