Metamath Proof Explorer


Theorem reusv2lem1

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

Ref Expression
Assertion reusv2lem1 ( 𝐴 ≠ ∅ → ( ∃! 𝑥𝑦𝐴 𝑥 = 𝐵 ↔ ∃ 𝑥𝑦𝐴 𝑥 = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 n0 ( 𝐴 ≠ ∅ ↔ ∃ 𝑦 𝑦𝐴 )
2 nfra1 𝑦𝑦𝐴 𝑥 = 𝐵
3 2 nfmov 𝑦 ∃* 𝑥𝑦𝐴 𝑥 = 𝐵
4 rsp ( ∀ 𝑦𝐴 𝑥 = 𝐵 → ( 𝑦𝐴𝑥 = 𝐵 ) )
5 4 com12 ( 𝑦𝐴 → ( ∀ 𝑦𝐴 𝑥 = 𝐵𝑥 = 𝐵 ) )
6 5 alrimiv ( 𝑦𝐴 → ∀ 𝑥 ( ∀ 𝑦𝐴 𝑥 = 𝐵𝑥 = 𝐵 ) )
7 mo2icl ( ∀ 𝑥 ( ∀ 𝑦𝐴 𝑥 = 𝐵𝑥 = 𝐵 ) → ∃* 𝑥𝑦𝐴 𝑥 = 𝐵 )
8 6 7 syl ( 𝑦𝐴 → ∃* 𝑥𝑦𝐴 𝑥 = 𝐵 )
9 3 8 exlimi ( ∃ 𝑦 𝑦𝐴 → ∃* 𝑥𝑦𝐴 𝑥 = 𝐵 )
10 1 9 sylbi ( 𝐴 ≠ ∅ → ∃* 𝑥𝑦𝐴 𝑥 = 𝐵 )
11 df-eu ( ∃! 𝑥𝑦𝐴 𝑥 = 𝐵 ↔ ( ∃ 𝑥𝑦𝐴 𝑥 = 𝐵 ∧ ∃* 𝑥𝑦𝐴 𝑥 = 𝐵 ) )
12 11 rbaib ( ∃* 𝑥𝑦𝐴 𝑥 = 𝐵 → ( ∃! 𝑥𝑦𝐴 𝑥 = 𝐵 ↔ ∃ 𝑥𝑦𝐴 𝑥 = 𝐵 ) )
13 10 12 syl ( 𝐴 ≠ ∅ → ( ∃! 𝑥𝑦𝐴 𝑥 = 𝐵 ↔ ∃ 𝑥𝑦𝐴 𝑥 = 𝐵 ) )