Theorem rexprg 3777

Description: Convert a quantification over a pair to a disjunction. (Contributed by NM, 17-Sep-2011.) (Revised by Mario Carneiro, 23-Apr-2015.)

Hypotheses

Ref	Expression
ralprg.1	⊢ (x = A → (φ ↔ ψ))
ralprg.2	⊢ (x = B → (φ ↔ χ))

Assertion

Ref	Expression
rexprg	⊢ ((A ∈ V ∧ B ∈ W) → (∃x ∈ {A, B}φ ↔ (ψ ∨ χ)))

Distinct variable groups:   x,A   x,B   ψ,x   χ,x

Allowed substitution hints:   φ(x)   V(x)   W(x)

Proof of Theorem rexprg

Step	Hyp	Ref	Expression
1		df-pr 3743	. . . 4 ⊢ {A, B} = ({A} ∪ {B})
2	1	rexeqi 2813	. . 3 ⊢ (∃x ∈ {A, B}φ ↔ ∃x ∈ ({A} ∪ {B})φ)
3		rexun 3444	. . 3 ⊢ (∃x ∈ ({A} ∪ {B})φ ↔ (∃x ∈ {A}φ ∨ ∃x ∈ {B}φ))
4	2, 3	bitri 240	. 2 ⊢ (∃x ∈ {A, B}φ ↔ (∃x ∈ {A}φ ∨ ∃x ∈ {B}φ))
5		ralprg.1	. . . . 5 ⊢ (x = A → (φ ↔ ψ))
6	5	rexsng 3767	. . . 4 ⊢ (A ∈ V → (∃x ∈ {A}φ ↔ ψ))
7	6	orbi1d 683	. . 3 ⊢ (A ∈ V → ((∃x ∈ {A}φ ∨ ∃x ∈ {B}φ) ↔ (ψ ∨ ∃x ∈ {B}φ)))
8		ralprg.2	. . . . 5 ⊢ (x = B → (φ ↔ χ))
9	8	rexsng 3767	. . . 4 ⊢ (B ∈ W → (∃x ∈ {B}φ ↔ χ))
10	9	orbi2d 682	. . 3 ⊢ (B ∈ W → ((ψ ∨ ∃x ∈ {B}φ) ↔ (ψ ∨ χ)))
11	7, 10	sylan9bb 680	. 2 ⊢ ((A ∈ V ∧ B ∈ W) → ((∃x ∈ {A}φ ∨ ∃x ∈ {B}φ) ↔ (ψ ∨ χ)))
12	4, 11	syl5bb 248	1 ⊢ ((A ∈ V ∧ B ∈ W) → (∃x ∈ {A, B}φ ↔ (ψ ∨ χ)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∨ wo 357   ∧ wa 358   = wceq 1642   ∈ wcel 1710  ∃wrex 2616   ∪ cun 3208  {csn 3738  {cpr 3739

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-3an 936  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-rex 2621  df-v 2862  df-sbc 3048  df-nin 3212  df-compl 3213  df-un 3215  df-sn 3742  df-pr 3743

This theorem is used by:  rextpg  3779  rexpr  3781