Theorem nelrdva 3727

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 2741	. 2 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 = 𝐵)
2		eleq1 2832	. . . . . . 7 ⊢ (𝑥 = 𝐵 → (𝑥 ∈ 𝐴 ↔ 𝐵 ∈ 𝐴))
3	2	anbi2d 629	. . . . . 6 ⊢ (𝑥 = 𝐵 → ((𝜑 ∧ 𝑥 ∈ 𝐴) ↔ (𝜑 ∧ 𝐵 ∈ 𝐴)))
4		neeq1 3009	. . . . . 6 ⊢ (𝑥 = 𝐵 → (𝑥 ≠ 𝐵 ↔ 𝐵 ≠ 𝐵))
5	3, 4	imbi12d 344	. . . . 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 3566	. . . 4 ⊢ (𝐵 ∈ 𝐴 → ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵))
8	7	anabsi7 670	. . 3 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵)
9	8	neneqd 2951	. 2 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → ¬ 𝐵 = 𝐵)
10	1, 9	pm2.65da 816	1 ⊢ (𝜑 → ¬ 𝐵 ∈ 𝐴)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395   = wceq 1537   ∈ wcel 2108   ≠ wne 2946

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ne 2947

This theorem is referenced by:  ustfilxp  24242  metustfbas  24591  drngmxidl  33470  fourierdlem72  46099