Theorem 2reu5lem1 3761

Description: Lemma for 2reu5 3764. 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 2656. (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,𝑦

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

Proof of Theorem 2reu5lem1

Step	Hyp	Ref	Expression
1		df-reu 3381	. . 3 ⊢ (∃!𝑦 ∈ 𝐵 𝜑 ↔ ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑))
2	1	reubii 3389	. 2 ⊢ (∃!𝑥 ∈ 𝐴 ∃!𝑦 ∈ 𝐵 𝜑 ↔ ∃!𝑥 ∈ 𝐴 ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑))
3		df-reu 3381	. . 3 ⊢ (∃!𝑥 ∈ 𝐴 ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑) ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)))
4		euanv 2624	. . . . . 6 ⊢ (∃!𝑦(𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐴 ∧ ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)))
5	4	bicomi 224	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 ∧ ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ ∃!𝑦(𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)))
6		3anass 1095	. . . . . . 7 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)))
7	6	bicomi 224	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
8	7	eubii 2585	. . . . 5 ⊢ (∃!𝑦(𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ ∃!𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
9	5, 8	bitri 275	. . . 4 ⊢ ((𝑥 ∈ 𝐴 ∧ ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ ∃!𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
10	9	eubii 2585	. . 3 ⊢ (∃!𝑥(𝑥 ∈ 𝐴 ∧ ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ ∃!𝑥∃!𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
11	3, 10	bitri 275	. 2 ⊢ (∃!𝑥 ∈ 𝐴 ∃!𝑦(𝑦 ∈ 𝐵 ∧ 𝜑) ↔ ∃!𝑥∃!𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
12	2, 11	bitri 275	1 ⊢ (∃!𝑥 ∈ 𝐴 ∃!𝑦 ∈ 𝐵 𝜑 ↔ ∃!𝑥∃!𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))

Syntax hints:   ↔ wb 206   ∧ wa 395   ∧ w3a 1087   ∈ wcel 2108  ∃!weu 2568  ∃!wreu 3378

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1089  df-ex 1780  df-mo 2540  df-eu 2569  df-reu 3381

This theorem is referenced by:  2reu5lem3  3763