Theorem nelrdva 3670

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 2766	. 2 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 = 𝐵)
2		eleq1 2853	. . . . . . 7 ⊢ (𝑥 = 𝐵 → (𝑥 ∈ 𝐴 ↔ 𝐵 ∈ 𝐴))
3	2	anbi2d 642	. . . . . 6 ⊢ (𝑥 = 𝐵 → ((𝜑 ∧ 𝑥 ∈ 𝐴) ↔ (𝜑 ∧ 𝐵 ∈ 𝐴)))
4		neeq1 3022	. . . . . 6 ⊢ (𝑥 = 𝐵 → (𝑥 ≠ 𝐵 ↔ 𝐵 ≠ 𝐵))
5	3, 4	imbi12d 347	. . . . 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 3524	. . . 4 ⊢ (𝐵 ∈ 𝐴 → ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵))
8	7	anabsi7 684	. . 3 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵)
9	8	neneqd 2965	. 2 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → ¬ 𝐵 = 𝐵)
10	1, 9	pm2.65da 829	1 ⊢ (𝜑 → ¬ 𝐵 ∈ 𝐴)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2146   ≠ wne 2960

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 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ne 2961

This theorem is used by:  ustfilxp  24423  metustfbas  24767  drngmxidl  33825  dflringlem2  33851  fourierdlem72  46952