Theorem sborv 1811

Description: Version of sbor 1869 where 𝑥 and 𝑦 are distinct. (Contributed by Jim Kingdon, 3-Feb-2018.)

Assertion

Ref	Expression
sborv	⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑥,𝑦)

Proof of Theorem sborv

Step	Hyp	Ref	Expression
1		sb5 1808	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ∃𝑥(𝑥 = 𝑦 ∧ (𝜑 ∨ 𝜓)))
2		andi 764	. . . 4 ⊢ ((𝑥 = 𝑦 ∧ (𝜑 ∨ 𝜓)) ↔ ((𝑥 = 𝑦 ∧ 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜓)))
3	2	exbii 1536	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ (𝜑 ∨ 𝜓)) ↔ ∃𝑥((𝑥 = 𝑦 ∧ 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜓)))
4		19.43 1559	. . 3 ⊢ (∃𝑥((𝑥 = 𝑦 ∧ 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜓)) ↔ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ∨ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓)))
5	1, 3, 4	3bitri 204	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ∨ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓)))
6		sb5 1808	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
7		sb5 1808	. . 3 ⊢ ([𝑦 / 𝑥]𝜓 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓))
8	6, 7	orbi12i 713	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓) ↔ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ∨ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓)))
9	5, 8	bitr4i 185	1 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))

Syntax hints:   ∧ wa 102   ↔ wb 103   ∨ wo 661  ∃wex 1421  [wsb 1685

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-11 1437  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467

This theorem depends on definitions:  df-bi 115  df-sb 1686

This theorem is referenced by:  sbor  1869