Theorem orbidi 955

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		pm5.74 270	. 2 ⊢ ((¬ 𝜑 → (𝜓 ↔ 𝜒)) ↔ ((¬ 𝜑 → 𝜓) ↔ (¬ 𝜑 → 𝜒)))
2		df-or 849	. 2 ⊢ ((𝜑 ∨ (𝜓 ↔ 𝜒)) ↔ (¬ 𝜑 → (𝜓 ↔ 𝜒)))
3		df-or 849	. . 3 ⊢ ((𝜑 ∨ 𝜓) ↔ (¬ 𝜑 → 𝜓))
4		df-or 849	. . 3 ⊢ ((𝜑 ∨ 𝜒) ↔ (¬ 𝜑 → 𝜒))
5	3, 4	bibi12i 339	. 2 ⊢ (((𝜑 ∨ 𝜓) ↔ (𝜑 ∨ 𝜒)) ↔ ((¬ 𝜑 → 𝜓) ↔ (¬ 𝜑 → 𝜒)))
6	1, 2, 5	3bitr4i 303	1 ⊢ ((𝜑 ∨ (𝜓 ↔ 𝜒)) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ∨ 𝜒)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ∨ wo 848

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 207  df-or 849

This theorem is referenced by:  pm5.7  956  dfclnbgr6  47842