Theorem ordir 453

Description: Distributive law for disjunction.

Assertion

Ref	Expression
ordir	⊢ (((φ ∧ ψ) ∨ χ) ↔ ((φ ∨ χ) ∧ (ψ ∨ χ)))

Proof of Theorem ordir

Step	Hyp	Ref	Expression
1		ordi 452	. 2 ⊢ ((χ ∨ (φ ∧ ψ)) ↔ ((χ ∨ φ) ∧ (χ ∨ ψ)))
2		orcom 209	. 2 ⊢ (((φ ∧ ψ) ∨ χ) ↔ (χ ∨ (φ ∧ ψ)))
3		orcom 209	. . 3 ⊢ ((φ ∨ χ) ↔ (χ ∨ φ))
4		orcom 209	. . 3 ⊢ ((ψ ∨ χ) ↔ (χ ∨ ψ))
5	3, 4	anbi12i 369	. 2 ⊢ (((φ ∨ χ) ∧ (ψ ∨ χ)) ↔ ((χ ∨ φ) ∧ (χ ∨ ψ)))
6	1, 2, 5	3bitr4 158	1 ⊢ (((φ ∧ ψ) ∨ χ) ↔ ((φ ∨ χ) ∧ (ψ ∨ χ)))

Syntax hints:   ↔ wb 127   ∨ wo 195   ∧ wa 196

This theorem is referenced by:  orddi 458  mapdom2 3389  elnn0z 4574

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198

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