Metamath Proof Explorer
Description: Version of 19.45 with a disjoint variable condition, requiring fewer
axioms. (Contributed by NM, 12-Mar-1993)
|
|
Ref |
Expression |
|
Assertion |
19.45v |
⊢ ( ∃ 𝑥 ( 𝜑 ∨ 𝜓 ) ↔ ( 𝜑 ∨ ∃ 𝑥 𝜓 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
19.43 |
⊢ ( ∃ 𝑥 ( 𝜑 ∨ 𝜓 ) ↔ ( ∃ 𝑥 𝜑 ∨ ∃ 𝑥 𝜓 ) ) |
2 |
|
19.9v |
⊢ ( ∃ 𝑥 𝜑 ↔ 𝜑 ) |
3 |
2
|
orbi1i |
⊢ ( ( ∃ 𝑥 𝜑 ∨ ∃ 𝑥 𝜓 ) ↔ ( 𝜑 ∨ ∃ 𝑥 𝜓 ) ) |
4 |
1 3
|
bitri |
⊢ ( ∃ 𝑥 ( 𝜑 ∨ 𝜓 ) ↔ ( 𝜑 ∨ ∃ 𝑥 𝜓 ) ) |