Theorem 2reu5lem1 3413

Description: Lemma for 2reu5 3416. 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 2557. (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 2919	. . 3 ⊢ (∃!𝑦 ∈ 𝐵 𝜑 ↔ ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑))
2	1	reubii 3128	. 2 ⊢ (∃!𝑥 ∈ 𝐴 ∃!𝑦 ∈ 𝐵 𝜑 ↔ ∃!𝑥 ∈ 𝐴 ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑))
3		df-reu 2919	. . 3 ⊢ (∃!𝑥 ∈ 𝐴 ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑) ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)))
4		euanv 2534	. . . . . 6 ⊢ (∃!𝑦(𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐴 ∧ ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)))
5	4	bicomi 214	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 ∧ ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ ∃!𝑦(𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)))
6		3anass 1042	. . . . . . 7 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)))
7	6	bicomi 214	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
8	7	eubii 2492	. . . . 5 ⊢ (∃!𝑦(𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ ∃!𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
9	5, 8	bitri 264	. . . 4 ⊢ ((𝑥 ∈ 𝐴 ∧ ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ ∃!𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
10	9	eubii 2492	. . 3 ⊢ (∃!𝑥(𝑥 ∈ 𝐴 ∧ ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ ∃!𝑥∃!𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
11	3, 10	bitri 264	. 2 ⊢ (∃!𝑥 ∈ 𝐴 ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑) ↔ ∃!𝑥∃!𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
12	2, 11	bitri 264	1 ⊢ (∃!𝑥 ∈ 𝐴 ∃!𝑦 ∈ 𝐵 𝜑 ↔ ∃!𝑥∃!𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))

Syntax hints:   ↔ wb 196   ∧ wa 384   ∧ w3a 1037   ∈ wcel 1990  ∃!weu 2470  ∃!wreu 2914

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-12 2047

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-ex 1705  df-nf 1710  df-eu 2474  df-reu 2919

This theorem is referenced by:  2reu5lem3  3415