Theorem nelrdva 3558

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 2761	. 2 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 = 𝐵)
2		eleq1 2827	. . . . . . 7 ⊢ (𝑥 = 𝐵 → (𝑥 ∈ 𝐴 ↔ 𝐵 ∈ 𝐴))
3	2	anbi2d 742	. . . . . 6 ⊢ (𝑥 = 𝐵 → ((𝜑 ∧ 𝑥 ∈ 𝐴) ↔ (𝜑 ∧ 𝐵 ∈ 𝐴)))
4		neeq1 2994	. . . . . 6 ⊢ (𝑥 = 𝐵 → (𝑥 ≠ 𝐵 ↔ 𝐵 ≠ 𝐵))
5	3, 4	imbi12d 333	. . . . 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 3406	. . . 4 ⊢ (𝐵 ∈ 𝐴 → ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵))
8	7	anabsi7 895	. . 3 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵)
9	8	neneqd 2937	. 2 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → ¬ 𝐵 = 𝐵)
10	1, 9	pm2.65da 601	1 ⊢ (𝜑 → ¬ 𝐵 ∈ 𝐴)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 383   = wceq 1632   ∈ wcel 2139   ≠ wne 2932

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090  ax-9 2148  ax-12 2196  ax-ext 2740

This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1635  df-ex 1854  df-nf 1859  df-sb 2047  df-clab 2747  df-cleq 2753  df-clel 2756  df-ne 2933  df-v 3342

This theorem is referenced by:  ustfilxp  22237  metustfbas  22583  fourierdlem72  40916