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 ) |
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 ) |