Theorem nelrdva 3640

Description: Deduce negative membership from an implication. (Contributed by Thierry Arnoux, 27-Nov-2017.)

Hypothesis

Ref	Expression
nelrdva.1	⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 ≠ 𝐵)

Assertion

Ref	Expression
nelrdva	⊢ (𝜑 → ¬ 𝐵 ∈ 𝐴)

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

Proof of Theorem nelrdva

Step	Hyp	Ref	Expression
1		eqidd 2739	. 2 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 = 𝐵)
2		eleq1 2826	. . . . . . 7 ⊢ (𝑥 = 𝐵 → (𝑥 ∈ 𝐴 ↔ 𝐵 ∈ 𝐴))
3	2	anbi2d 629	. . . . . 6 ⊢ (𝑥 = 𝐵 → ((𝜑 ∧ 𝑥 ∈ 𝐴) ↔ (𝜑 ∧ 𝐵 ∈ 𝐴)))
4		neeq1 3006	. . . . . 6 ⊢ (𝑥 = 𝐵 → (𝑥 ≠ 𝐵 ↔ 𝐵 ≠ 𝐵))
5	3, 4	imbi12d 345	. . . . 5 ⊢ (𝑥 = 𝐵 → (((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 ≠ 𝐵) ↔ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵)))
6		nelrdva.1	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 ≠ 𝐵)
7	5, 6	vtoclg 3505	. . . 4 ⊢ (𝐵 ∈ 𝐴 → ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵))
8	7	anabsi7 668	. . 3 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵)
9	8	neneqd 2948	. 2 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → ¬ 𝐵 = 𝐵)
10	1, 9	pm2.65da 814	1 ⊢ (𝜑 → ¬ 𝐵 ∈ 𝐴)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 396   = wceq 1539   ∈ wcel 2106   ≠ wne 2943

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ne 2944

This theorem is referenced by:  ustfilxp  23364  metustfbas  23713  fourierdlem72  43719