Theorem orddi 458

Description: Double distributive law for disjunction.

Assertion

Ref	Expression
orddi	⊢ (((φ ∧ ψ) ∨ (χ ∧ θ)) ↔ (((φ ∨ χ) ∧ (φ ∨ θ)) ∧ ((ψ ∨ χ) ∧ (ψ ∨ θ))))

Proof of Theorem orddi

Step	Hyp	Ref	Expression
1		ordir 453	. 2 ⊢ (((φ ∧ ψ) ∨ (χ ∧ θ)) ↔ ((φ ∨ (χ ∧ θ)) ∧ (ψ ∨ (χ ∧ θ))))
2		ordi 452	. . 3 ⊢ ((φ ∨ (χ ∧ θ)) ↔ ((φ ∨ χ) ∧ (φ ∨ θ)))
3		ordi 452	. . 3 ⊢ ((ψ ∨ (χ ∧ θ)) ↔ ((ψ ∨ χ) ∧ (ψ ∨ θ)))
4	2, 3	anbi12i 369	. 2 ⊢ (((φ ∨ (χ ∧ θ)) ∧ (ψ ∨ (χ ∧ θ))) ↔ (((φ ∨ χ) ∧ (φ ∨ θ)) ∧ ((ψ ∨ χ) ∧ (ψ ∨ θ))))
5	1, 4	bitr 151	1 ⊢ (((φ ∧ ψ) ∨ (χ ∧ θ)) ↔ (((φ ∨ χ) ∧ (φ ∨ θ)) ∧ ((ψ ∨ χ) ∧ (ψ ∨ θ))))

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

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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