Theorem reusv2lem1 5360

Description: Lemma for reusv2 5365. (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 4300	. . 3 ⊢ (𝐴 ≠ ∅ ↔ ∃𝑦 𝑦 ∈ 𝐴)
2		nfra1 3287	. . . . 5 ⊢ Ⅎ𝑦∀𝑦 ∈ 𝐴 𝑥 = 𝐵
3	2	nfmov 2586	. . . 4 ⊢ Ⅎ𝑦∃*𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵
4		rsp 3251	. . . . . . 7 ⊢ (∀𝑦 ∈ 𝐴 𝑥 = 𝐵 → (𝑦 ∈ 𝐴 → 𝑥 = 𝐵))
5	4	com12 33	. . . . . 6 ⊢ (𝑦 ∈ 𝐴 → (∀𝑦 ∈ 𝐴 𝑥 = 𝐵 → 𝑥 = 𝐵))
6	5	alrimiv 1960	. . . . 5 ⊢ (𝑦 ∈ 𝐴 → ∀𝑥(∀𝑦 ∈ 𝐴 𝑥 = 𝐵 → 𝑥 = 𝐵))
7		mo2icl 3672	. . . . 5 ⊢ (∀𝑥(∀𝑦 ∈ 𝐴 𝑥 = 𝐵 → 𝑥 = 𝐵) → ∃*𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵)
8	6, 7	syl 18	. . . 4 ⊢ (𝑦 ∈ 𝐴 → ∃*𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵)
9	3, 8	exlimi 2254	. . 3 ⊢ (∃𝑦 𝑦 ∈ 𝐴 → ∃*𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵)
10	1, 9	sylbi 220	. 2 ⊢ (𝐴 ≠ ∅ → ∃*𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵)
11		df-eu 2595	. . 3 ⊢ (∃!𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵 ↔ (∃𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵 ∧ ∃*𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵))
12	11	rbaib 548	. 2 ⊢ (∃*𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵 → (∃!𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵 ↔ ∃𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵))
13	10, 12	syl 18	1 ⊢ (𝐴 ≠ ∅ → (∃!𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵 ↔ ∃𝑥∀𝑦 ∈ 𝐴 𝑥 = 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568   = wceq 1570  ∃wex 1812   ∈ wcel 2145  ∃*wmo 2563  ∃!weu 2594   ≠ wne 2956  ∀wral 3077  ∅c0 4279

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-mo 2565  df-eu 2595  df-clab 2740  df-cleq 2753  df-clel 2836  df-ne 2957  df-ral 3078  df-v 3453  df-dif 3902  df-nul 4280

This theorem is used by: (None)