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Description: Deduce equality of classes from an equivalence of membership that depends on the membership variable. (Contributed by NM, 7-Nov-2008) (Proof shortened by Wolf Lammen, 19-Nov-2019)
| Ref | Expression | ||
|---|---|---|---|
| Hypotheses | eqrdav.1 | ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝑥 ∈ 𝐶 ) | |
| eqrdav.2 | ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐵 ) → 𝑥 ∈ 𝐶 ) | ||
| eqrdav.3 | ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐶 ) → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) ) | ||
| Assertion | eqrdav | ⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | eqrdav.1 | ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝑥 ∈ 𝐶 ) | |
| 2 | eqrdav.2 | ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐵 ) → 𝑥 ∈ 𝐶 ) | |
| 3 | eqrdav.3 | ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐶 ) → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) ) | |
| 4 | 1 2 3 | bibiad | ⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) ) |
| 5 | 4 | eqrdv | ⊢ ( 𝜑 → 𝐴 = 𝐵 ) |