Theorem reusv2lem1 5398

Description: Lemma for reusv2 5403. (Contributed by NM, 22-Oct-2010.) (Proof shortened by Mario Carneiro, 19-Nov-2016.)

Assertion

Ref	Expression
reusv2lem1	⊢ (𝐴 ≠ ∅ → (∃!𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵 ↔ ∃𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵))

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

Allowed substitution hint:   𝐵(𝑦)

Proof of Theorem reusv2lem1

Step	Hyp	Ref	Expression
1		n0 4353	. . 3 ⊢ (𝐴 ≠ ∅ ↔ ∃𝑦 𝑦 ∈ 𝐴)
2		nfra1 3284	. . . . 5 ⊢ Ⅎ𝑦∀𝑦 ∈ 𝐴 𝑥 = 𝐵
3	2	nfmov 2560	. . . 4 ⊢ Ⅎ𝑦∃*𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵
4		rsp 3247	. . . . . . 7 ⊢ (∀𝑦 ∈ 𝐴 𝑥 = 𝐵 → (𝑦 ∈ 𝐴 → 𝑥 = 𝐵))
5	4	com12 32	. . . . . 6 ⊢ (𝑦 ∈ 𝐴 → (∀𝑦 ∈ 𝐴 𝑥 = 𝐵 → 𝑥 = 𝐵))
6	5	alrimiv 1927	. . . . 5 ⊢ (𝑦 ∈ 𝐴 → ∀𝑥(∀𝑦 ∈ 𝐴 𝑥 = 𝐵 → 𝑥 = 𝐵))
7		mo2icl 3720	. . . . 5 ⊢ (∀𝑥(∀𝑦 ∈ 𝐴 𝑥 = 𝐵 → 𝑥 = 𝐵) → ∃*𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵)
8	6, 7	syl 17	. . . 4 ⊢ (𝑦 ∈ 𝐴 → ∃*𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵)
9	3, 8	exlimi 2217	. . 3 ⊢ (∃𝑦 𝑦 ∈ 𝐴 → ∃*𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵)
10	1, 9	sylbi 217	. 2 ⊢ (𝐴 ≠ ∅ → ∃*𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵)
11		df-eu 2569	. . 3 ⊢ (∃!𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵 ↔ (∃𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵 ∧ ∃*𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵))
12	11	rbaib 538	. 2 ⊢ (∃*𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵 → (∃!𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵 ↔ ∃𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵))
13	10, 12	syl 17	1 ⊢ (𝐴 ≠ ∅ → (∃!𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵 ↔ ∃𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵))

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1538   = wceq 1540  ∃wex 1779   ∈ wcel 2108  ∃*wmo 2538  ∃!weu 2568   ≠ wne 2940  ∀wral 3061  ∅c0 4333

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  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2065  df-mo 2540  df-eu 2569  df-clab 2715  df-cleq 2729  df-clel 2816  df-ne 2941  df-ral 3062  df-v 3482  df-dif 3954  df-nul 4334

This theorem is referenced by: (None)