Theorem orbidi 898

Description: Disjunction distributes over the biconditional. An axiom of system DS in Vladimir Lifschitz, "On calculational proofs" (1998), http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.25.3384. (Contributed by NM, 8-Jan-2005.) (Proof shortened by Wolf Lammen, 4-Feb-2013.)

Assertion

Ref	Expression
orbidi	⊢ ((φ ∨ (ψ ↔ χ)) ↔ ((φ ∨ ψ) ↔ (φ ∨ χ)))

Proof of Theorem orbidi

Step	Hyp	Ref	Expression
1		pm5.74 235	. 2 ⊢ ((¬ φ → (ψ ↔ χ)) ↔ ((¬ φ → ψ) ↔ (¬ φ → χ)))
2		df-or 359	. 2 ⊢ ((φ ∨ (ψ ↔ χ)) ↔ (¬ φ → (ψ ↔ χ)))
3		df-or 359	. . 3 ⊢ ((φ ∨ ψ) ↔ (¬ φ → ψ))
4		df-or 359	. . 3 ⊢ ((φ ∨ χ) ↔ (¬ φ → χ))
5	3, 4	bibi12i 306	. 2 ⊢ (((φ ∨ ψ) ↔ (φ ∨ χ)) ↔ ((¬ φ → ψ) ↔ (¬ φ → χ)))
6	1, 2, 5	3bitr4i 268	1 ⊢ ((φ ∨ (ψ ↔ χ)) ↔ ((φ ∨ ψ) ↔ (φ ∨ χ)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176   ∨ wo 357

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-or 359

This theorem is used by:  pm5.7  900