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Metamath Proof Explorer
Description: Version of 19.27 with a disjoint variable condition, requiring fewer
axioms. (Contributed by NM, 3-Jun-2004)
|
|
Ref |
Expression |
|
Assertion |
19.27v |
⊢ ( ∀ 𝑥 ( 𝜑 ∧ 𝜓 ) ↔ ( ∀ 𝑥 𝜑 ∧ 𝜓 ) ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
19.26 |
⊢ ( ∀ 𝑥 ( 𝜑 ∧ 𝜓 ) ↔ ( ∀ 𝑥 𝜑 ∧ ∀ 𝑥 𝜓 ) ) |
| 2 |
|
19.3v |
⊢ ( ∀ 𝑥 𝜓 ↔ 𝜓 ) |
| 3 |
2
|
anbi2i |
⊢ ( ( ∀ 𝑥 𝜑 ∧ ∀ 𝑥 𝜓 ) ↔ ( ∀ 𝑥 𝜑 ∧ 𝜓 ) ) |
| 4 |
1 3
|
bitri |
⊢ ( ∀ 𝑥 ( 𝜑 ∧ 𝜓 ) ↔ ( ∀ 𝑥 𝜑 ∧ 𝜓 ) ) |