Description: Obsolete version of rexeqf as of 9-Mar-2025. (Contributed by NM, 9-Oct-2003) (Revised by Andrew Salmon, 11-Jul-2011) (Proof modification is discouraged.) (New usage is discouraged.)
| Ref | Expression | ||
|---|---|---|---|
| Hypotheses | raleqf.1 | ⊢ Ⅎ 𝑥 𝐴 | |
| raleqf.2 | ⊢ Ⅎ 𝑥 𝐵 | ||
| Assertion | rexeqfOLD | ⊢ ( 𝐴 = 𝐵 → ( ∃ 𝑥 ∈ 𝐴 𝜑 ↔ ∃ 𝑥 ∈ 𝐵 𝜑 ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | raleqf.1 | ⊢ Ⅎ 𝑥 𝐴 | |
| 2 | raleqf.2 | ⊢ Ⅎ 𝑥 𝐵 | |
| 3 | 1 2 | nfeq | ⊢ Ⅎ 𝑥 𝐴 = 𝐵 |
| 4 | eleq2 | ⊢ ( 𝐴 = 𝐵 → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) ) | |
| 5 | 4 | anbi1d | ⊢ ( 𝐴 = 𝐵 → ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ↔ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) ) |
| 6 | 3 5 | exbid | ⊢ ( 𝐴 = 𝐵 → ( ∃ 𝑥 ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ↔ ∃ 𝑥 ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) ) |
| 7 | df-rex | ⊢ ( ∃ 𝑥 ∈ 𝐴 𝜑 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) | |
| 8 | df-rex | ⊢ ( ∃ 𝑥 ∈ 𝐵 𝜑 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) | |
| 9 | 6 7 8 | 3bitr4g | ⊢ ( 𝐴 = 𝐵 → ( ∃ 𝑥 ∈ 𝐴 𝜑 ↔ ∃ 𝑥 ∈ 𝐵 𝜑 ) ) |