Theorem orbididc 955

Description: Disjunction distributes over the biconditional, for a decidable proposition. Based on 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 Jim Kingdon, 2-Apr-2018.)

Assertion

Ref	Expression
orbididc	⊢ (DECID 𝜑 → ((𝜑 ∨ (𝜓 ↔ 𝜒)) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ∨ 𝜒))))

Proof of Theorem orbididc

Step	Hyp	Ref	Expression
1		orimdidc 907	. . 3 ⊢ (DECID 𝜑 → ((𝜑 ∨ (𝜓 → 𝜒)) ↔ ((𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜒))))
2		orimdidc 907	. . 3 ⊢ (DECID 𝜑 → ((𝜑 ∨ (𝜒 → 𝜓)) ↔ ((𝜑 ∨ 𝜒) → (𝜑 ∨ 𝜓))))
3	1, 2	anbi12d 473	. 2 ⊢ (DECID 𝜑 → (((𝜑 ∨ (𝜓 → 𝜒)) ∧ (𝜑 ∨ (𝜒 → 𝜓))) ↔ (((𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜒)) ∧ ((𝜑 ∨ 𝜒) → (𝜑 ∨ 𝜓)))))
4		dfbi2 388	. . . 4 ⊢ ((𝜓 ↔ 𝜒) ↔ ((𝜓 → 𝜒) ∧ (𝜒 → 𝜓)))
5	4	orbi2i 763	. . 3 ⊢ ((𝜑 ∨ (𝜓 ↔ 𝜒)) ↔ (𝜑 ∨ ((𝜓 → 𝜒) ∧ (𝜒 → 𝜓))))
6		ordi 817	. . 3 ⊢ ((𝜑 ∨ ((𝜓 → 𝜒) ∧ (𝜒 → 𝜓))) ↔ ((𝜑 ∨ (𝜓 → 𝜒)) ∧ (𝜑 ∨ (𝜒 → 𝜓))))
7	5, 6	bitri 184	. 2 ⊢ ((𝜑 ∨ (𝜓 ↔ 𝜒)) ↔ ((𝜑 ∨ (𝜓 → 𝜒)) ∧ (𝜑 ∨ (𝜒 → 𝜓))))
8		dfbi2 388	. 2 ⊢ (((𝜑 ∨ 𝜓) ↔ (𝜑 ∨ 𝜒)) ↔ (((𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜒)) ∧ ((𝜑 ∨ 𝜒) → (𝜑 ∨ 𝜓))))
9	3, 7, 8	3bitr4g 223	1 ⊢ (DECID 𝜑 → ((𝜑 ∨ (𝜓 ↔ 𝜒)) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ∨ 𝜒))))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   ∨ wo 709  DECID wdc 835

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in2 616  ax-io 710

This theorem depends on definitions:  df-bi 117  df-dc 836

This theorem is referenced by:  pm5.7dc  956