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Metamath Proof Explorer
Description: Version of 19.37 with a disjoint variable condition, requiring fewer
axioms. (Contributed by NM, 21-Jun-1993)
|
|
Ref |
Expression |
|
Assertion |
19.37v |
⊢ ( ∃ 𝑥 ( 𝜑 → 𝜓 ) ↔ ( 𝜑 → ∃ 𝑥 𝜓 ) ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
19.35 |
⊢ ( ∃ 𝑥 ( 𝜑 → 𝜓 ) ↔ ( ∀ 𝑥 𝜑 → ∃ 𝑥 𝜓 ) ) |
| 2 |
|
19.3v |
⊢ ( ∀ 𝑥 𝜑 ↔ 𝜑 ) |
| 3 |
2
|
imbi1i |
⊢ ( ( ∀ 𝑥 𝜑 → ∃ 𝑥 𝜓 ) ↔ ( 𝜑 → ∃ 𝑥 𝜓 ) ) |
| 4 |
1 3
|
bitri |
⊢ ( ∃ 𝑥 ( 𝜑 → 𝜓 ) ↔ ( 𝜑 → ∃ 𝑥 𝜓 ) ) |