Metamath Proof Explorer


Theorem nelrdva

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

Ref Expression
Hypothesis nelrdva.1 ⊢ φ ∧ x ∈ A → x ≠ B
Assertion nelrdva ⊢ φ → ¬ B ∈ A

Proof

Step Hyp Ref Expression
1 nelrdva.1 ⊢ φ ∧ x ∈ A → x ≠ B
2 eqidd ⊢ φ ∧ B ∈ A → B = B
3 eleq1 ⊢ x = B → x ∈ A ↔ B ∈ A
4 3 anbi2d ⊢ x = B → φ ∧ x ∈ A ↔ φ ∧ B ∈ A
5 neeq1 ⊢ x = B → x ≠ B ↔ B ≠ B
6 4 5 imbi12d ⊢ x = B → φ ∧ x ∈ A → x ≠ B ↔ φ ∧ B ∈ A → B ≠ B
7 6 1 vtoclg ⊢ B ∈ A → φ ∧ B ∈ A → B ≠ B
8 7 anabsi7 ⊢ φ ∧ B ∈ A → B ≠ B
9 8 neneqd ⊢ φ ∧ B ∈ A → ¬ B = B
10 2 9 pm2.65da ⊢ φ → ¬ B ∈ A