Metamath Proof Explorer
Description: Double application of rspcedvdw . (Contributed by SN, 24-Aug-2024)
|
|
Ref |
Expression |
|
Hypotheses |
2rspcedvdw.1 |
⊢ ( 𝑥 = 𝐴 → ( 𝜓 ↔ 𝜒 ) ) |
|
|
2rspcedvdw.2 |
⊢ ( 𝑦 = 𝐵 → ( 𝜒 ↔ 𝜃 ) ) |
|
|
2rspcedvdw.a |
⊢ ( 𝜑 → 𝐴 ∈ 𝑋 ) |
|
|
2rspcedvdw.b |
⊢ ( 𝜑 → 𝐵 ∈ 𝑌 ) |
|
|
2rspcedvdw.3 |
⊢ ( 𝜑 → 𝜃 ) |
|
Assertion |
2rspcedvdw |
⊢ ( 𝜑 → ∃ 𝑥 ∈ 𝑋 ∃ 𝑦 ∈ 𝑌 𝜓 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
2rspcedvdw.1 |
⊢ ( 𝑥 = 𝐴 → ( 𝜓 ↔ 𝜒 ) ) |
2 |
|
2rspcedvdw.2 |
⊢ ( 𝑦 = 𝐵 → ( 𝜒 ↔ 𝜃 ) ) |
3 |
|
2rspcedvdw.a |
⊢ ( 𝜑 → 𝐴 ∈ 𝑋 ) |
4 |
|
2rspcedvdw.b |
⊢ ( 𝜑 → 𝐵 ∈ 𝑌 ) |
5 |
|
2rspcedvdw.3 |
⊢ ( 𝜑 → 𝜃 ) |
6 |
1 2
|
rspc2ev |
⊢ ( ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑌 ∧ 𝜃 ) → ∃ 𝑥 ∈ 𝑋 ∃ 𝑦 ∈ 𝑌 𝜓 ) |
7 |
3 4 5 6
|
syl3anc |
⊢ ( 𝜑 → ∃ 𝑥 ∈ 𝑋 ∃ 𝑦 ∈ 𝑌 𝜓 ) |