Theorem sborv 1878

Description: Version of sbor 1942 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 1875	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ∃𝑥(𝑥 = 𝑦 ∧ (𝜑 ∨ 𝜓)))
2		andi 808	. . . 4 ⊢ ((𝑥 = 𝑦 ∧ (𝜑 ∨ 𝜓)) ↔ ((𝑥 = 𝑦 ∧ 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜓)))
3	2	exbii 1593	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ (𝜑 ∨ 𝜓)) ↔ ∃𝑥((𝑥 = 𝑦 ∧ 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜓)))
4		19.43 1616	. . 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 1875	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
7		sb5 1875	. . 3 ⊢ ([𝑦 / 𝑥]𝜓 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓))
8	6, 7	orbi12i 754	. 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 698  ∃wex 1480  [wsb 1750

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 699  ax-5 1435  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-11 1494  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522

This theorem depends on definitions:  df-bi 116  df-sb 1751

This theorem is referenced by:  sbor  1942