Description: Obsolete version of rabid2 as of 24-11-2024. (Contributed by NM, 9-Oct-2003) (Proof shortened by Andrew Salmon, 30-May-2011) (Proof modification is discouraged.) (New usage is discouraged.)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | rabid2OLD | ⊢ ( 𝐴 = { 𝑥 ∈ 𝐴 ∣ 𝜑 } ↔ ∀ 𝑥 ∈ 𝐴 𝜑 ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | abeq2 | ⊢ ( 𝐴 = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ) | |
| 2 | pm4.71 | ⊢ ( ( 𝑥 ∈ 𝐴 → 𝜑 ) ↔ ( 𝑥 ∈ 𝐴 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ) | |
| 3 | 2 | albii | ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜑 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ) |
| 4 | 1 3 | bitr4i | ⊢ ( 𝐴 = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜑 ) ) |
| 5 | df-rab | ⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } | |
| 6 | 5 | eqeq2i | ⊢ ( 𝐴 = { 𝑥 ∈ 𝐴 ∣ 𝜑 } ↔ 𝐴 = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ) |
| 7 | df-ral | ⊢ ( ∀ 𝑥 ∈ 𝐴 𝜑 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜑 ) ) | |
| 8 | 4 6 7 | 3bitr4i | ⊢ ( 𝐴 = { 𝑥 ∈ 𝐴 ∣ 𝜑 } ↔ ∀ 𝑥 ∈ 𝐴 𝜑 ) |