Description: Obsolete version of rabeqbidva as of 1-Sep-2025. (Contributed by Mario Carneiro, 26-Jan-2017) (Proof modification is discouraged.) (New usage is discouraged.)
Ref | Expression | ||
---|---|---|---|
Hypotheses | rabeqbidvaOLD.1 | ⊢ ( 𝜑 → 𝐴 = 𝐵 ) | |
rabeqbidvaOLD.2 | ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝜓 ↔ 𝜒 ) ) | ||
Assertion | rabeqbidvaOLD | ⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜒 } ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | rabeqbidvaOLD.1 | ⊢ ( 𝜑 → 𝐴 = 𝐵 ) | |
2 | rabeqbidvaOLD.2 | ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝜓 ↔ 𝜒 ) ) | |
3 | 2 | rabbidva | ⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐴 ∣ 𝜒 } ) |
4 | 1 | rabeqdv | ⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜒 } = { 𝑥 ∈ 𝐵 ∣ 𝜒 } ) |
5 | 3 4 | eqtrd | ⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜒 } ) |