Description: Deduction form of rabeq . Note that contrary to rabeq it has no disjoint variable condition. (Contributed by BJ, 27-Apr-2019)
Ref | Expression | ||
---|---|---|---|
Hypotheses | bj-rabeqd.nf | ⊢ Ⅎ 𝑥 𝜑 | |
bj-rabeqd.1 | ⊢ ( 𝜑 → 𝐴 = 𝐵 ) | ||
Assertion | bj-rabeqd | ⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜓 } ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | bj-rabeqd.nf | ⊢ Ⅎ 𝑥 𝜑 | |
2 | bj-rabeqd.1 | ⊢ ( 𝜑 → 𝐴 = 𝐵 ) | |
3 | eleq2 | ⊢ ( 𝐴 = 𝐵 → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) ) | |
4 | 3 | anbi1d | ⊢ ( 𝐴 = 𝐵 → ( ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ↔ ( 𝑥 ∈ 𝐵 ∧ 𝜓 ) ) ) |
5 | 2 4 | syl | ⊢ ( 𝜑 → ( ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ↔ ( 𝑥 ∈ 𝐵 ∧ 𝜓 ) ) ) |
6 | 1 5 | bj-rabbida2 | ⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜓 } ) |