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Description: Alternate proof of ab0 , shorter but using more axioms. (Contributed by BJ, 19-Mar-2021) (Proof modification is discouraged.) (New usage is discouraged.)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | ab0ALT | ⊢ ( { 𝑥 ∣ 𝜑 } = ∅ ↔ ∀ 𝑥 ¬ 𝜑 ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | nfab1 | ⊢ Ⅎ 𝑥 { 𝑥 ∣ 𝜑 } | |
| 2 | 1 | eq0f | ⊢ ( { 𝑥 ∣ 𝜑 } = ∅ ↔ ∀ 𝑥 ¬ 𝑥 ∈ { 𝑥 ∣ 𝜑 } ) |
| 3 | abid | ⊢ ( 𝑥 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜑 ) | |
| 4 | 3 | notbii | ⊢ ( ¬ 𝑥 ∈ { 𝑥 ∣ 𝜑 } ↔ ¬ 𝜑 ) |
| 5 | 4 | albii | ⊢ ( ∀ 𝑥 ¬ 𝑥 ∈ { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ¬ 𝜑 ) |
| 6 | 2 5 | bitri | ⊢ ( { 𝑥 ∣ 𝜑 } = ∅ ↔ ∀ 𝑥 ¬ 𝜑 ) |