Theorem ralprg 3776

Description: Convert a quantification over a pair to a conjunction. (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
ralprg	⊢ ((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 ralprg

Step	Hyp	Ref	Expression
1		df-pr 3743	. . . 4 ⊢ {A, B} = ({A} ∪ {B})
2	1	raleqi 2812	. . 3 ⊢ (∀x ∈ {A, B}φ ↔ ∀x ∈ ({A} ∪ {B})φ)
3		ralunb 3445	. . 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	. . . 4 ⊢ (x = A → (φ ↔ ψ))
6	5	ralsng 3766	. . 3 ⊢ (A ∈ V → (∀x ∈ {A}φ ↔ ψ))
7		ralprg.2	. . . 4 ⊢ (x = B → (φ ↔ χ))
8	7	ralsng 3766	. . 3 ⊢ (B ∈ W → (∀x ∈ {B}φ ↔ χ))
9	6, 8	bi2anan9 843	. 2 ⊢ ((A ∈ V ∧ B ∈ W) → ((∀x ∈ {A}φ ∧ ∀x ∈ {B}φ) ↔ (ψ ∧ χ)))
10	4, 9	syl5bb 248	1 ⊢ ((A ∈ V ∧ B ∈ W) → (∀x ∈ {A, B}φ ↔ (ψ ∧ χ)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358   = wceq 1642   ∈ wcel 1710  ∀wral 2615   ∪ 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-ral 2620  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:  raltpg  3778  ralpr  3780  iinxprg  4044