Theorem sborv 1862

Description: Version of sbor 1927 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 1859	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ∃𝑥(𝑥 = 𝑦 ∧ (𝜑 ∨ 𝜓)))
2		andi 807	. . . 4 ⊢ ((𝑥 = 𝑦 ∧ (𝜑 ∨ 𝜓)) ↔ ((𝑥 = 𝑦 ∧ 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜓)))
3	2	exbii 1584	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ (𝜑 ∨ 𝜓)) ↔ ∃𝑥((𝑥 = 𝑦 ∧ 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜓)))
4		19.43 1607	. . 3 ⊢ (∃𝑥((𝑥 = 𝑦 ∧ 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜓)) ↔ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ∨ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓)))
5	1, 3, 4	3bitri 205	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ∨ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓)))
6		sb5 1859	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
7		sb5 1859	. . 3 ⊢ ([𝑦 / 𝑥]𝜓 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓))
8	6, 7	orbi12i 753	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓) ↔ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ∨ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓)))
9	5, 8	bitr4i 186	1 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))

Syntax hints:   ∧ wa 103   ↔ wb 104   ∨ wo 697  ∃wex 1468  [wsb 1735

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-11 1484  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514

This theorem depends on definitions:  df-bi 116  df-sb 1736

This theorem is referenced by:  sbor  1927