Theorem 2reu5lem1 3041

Description: Lemma for 2reu5 3044. Note that ∃!x ∈ A∃!y ∈ Bφ does not mean "there is exactly one x in A and exactly one y in B such that φ holds;" see comment for 2eu5 2288. (Contributed by Alexander van der Vekens, 17-Jun-2017.)

Assertion

Ref	Expression
2reu5lem1	⊢ (∃!x ∈ A ∃!y ∈ B φ ↔ ∃!x∃!y(x ∈ A ∧ y ∈ B ∧ φ))

Distinct variable groups:   y,A   x,B   x,y

Allowed substitution hints:   φ(x,y)   A(x)   B(y)

Proof of Theorem 2reu5lem1

Step	Hyp	Ref	Expression
1		df-reu 2621	. . 3 ⊢ (∃!y ∈ B φ ↔ ∃!y(y ∈ B ∧ φ))
2	1	reubii 2797	. 2 ⊢ (∃!x ∈ A ∃!y ∈ B φ ↔ ∃!x ∈ A ∃!y(y ∈ B ∧ φ))
3		df-reu 2621	. . 3 ⊢ (∃!x ∈ A ∃!y(y ∈ B ∧ φ) ↔ ∃!x(x ∈ A ∧ ∃!y(y ∈ B ∧ φ)))
4		euanv 2265	. . . . . 6 ⊢ (∃!y(x ∈ A ∧ (y ∈ B ∧ φ)) ↔ (x ∈ A ∧ ∃!y(y ∈ B ∧ φ)))
5	4	bicomi 193	. . . . 5 ⊢ ((x ∈ A ∧ ∃!y(y ∈ B ∧ φ)) ↔ ∃!y(x ∈ A ∧ (y ∈ B ∧ φ)))
6		3anass 938	. . . . . . 7 ⊢ ((x ∈ A ∧ y ∈ B ∧ φ) ↔ (x ∈ A ∧ (y ∈ B ∧ φ)))
7	6	bicomi 193	. . . . . 6 ⊢ ((x ∈ A ∧ (y ∈ B ∧ φ)) ↔ (x ∈ A ∧ y ∈ B ∧ φ))
8	7	eubii 2213	. . . . 5 ⊢ (∃!y(x ∈ A ∧ (y ∈ B ∧ φ)) ↔ ∃!y(x ∈ A ∧ y ∈ B ∧ φ))
9	5, 8	bitri 240	. . . 4 ⊢ ((x ∈ A ∧ ∃!y(y ∈ B ∧ φ)) ↔ ∃!y(x ∈ A ∧ y ∈ B ∧ φ))
10	9	eubii 2213	. . 3 ⊢ (∃!x(x ∈ A ∧ ∃!y(y ∈ B ∧ φ)) ↔ ∃!x∃!y(x ∈ A ∧ y ∈ B ∧ φ))
11	3, 10	bitri 240	. 2 ⊢ (∃!x ∈ A ∃!y(y ∈ B ∧ φ) ↔ ∃!x∃!y(x ∈ A ∧ y ∈ B ∧ φ))
12	2, 11	bitri 240	1 ⊢ (∃!x ∈ A ∃!y ∈ B φ ↔ ∃!x∃!y(x ∈ A ∧ y ∈ B ∧ φ))

Syntax hints:   ↔ wb 176   ∧ wa 358   ∧ w3a 934   ∈ wcel 1710  ∃!weu 2204  ∃!wreu 2616

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 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

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-3an 936  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-eu 2208  df-mo 2209  df-reu 2621

This theorem is referenced by:  2reu5lem3  3043