Description: Two ways of saying "the complement of a class abstraction". (Contributed by Andrew Salmon, 15-Jul-2011) (Proof shortened by Mario Carneiro, 11-Dec-2016)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | compab | ⊢ ( V ∖ { 𝑧 ∣ 𝜑 } ) = { 𝑧 ∣ ¬ 𝜑 } | 
| Step | Hyp | Ref | Expression | 
|---|---|---|---|
| 1 | nfcv | ⊢ Ⅎ 𝑧 V | |
| 2 | nfab1 | ⊢ Ⅎ 𝑧 { 𝑧 ∣ 𝜑 } | |
| 3 | 1 2 | nfdif | ⊢ Ⅎ 𝑧 ( V ∖ { 𝑧 ∣ 𝜑 } ) | 
| 4 | nfab1 | ⊢ Ⅎ 𝑧 { 𝑧 ∣ ¬ 𝜑 } | |
| 5 | 3 4 | cleqf | ⊢ ( ( V ∖ { 𝑧 ∣ 𝜑 } ) = { 𝑧 ∣ ¬ 𝜑 } ↔ ∀ 𝑧 ( 𝑧 ∈ ( V ∖ { 𝑧 ∣ 𝜑 } ) ↔ 𝑧 ∈ { 𝑧 ∣ ¬ 𝜑 } ) ) | 
| 6 | abid | ⊢ ( 𝑧 ∈ { 𝑧 ∣ 𝜑 } ↔ 𝜑 ) | |
| 7 | 6 | notbii | ⊢ ( ¬ 𝑧 ∈ { 𝑧 ∣ 𝜑 } ↔ ¬ 𝜑 ) | 
| 8 | velcomp | ⊢ ( 𝑧 ∈ ( V ∖ { 𝑧 ∣ 𝜑 } ) ↔ ¬ 𝑧 ∈ { 𝑧 ∣ 𝜑 } ) | |
| 9 | abid | ⊢ ( 𝑧 ∈ { 𝑧 ∣ ¬ 𝜑 } ↔ ¬ 𝜑 ) | |
| 10 | 7 8 9 | 3bitr4i | ⊢ ( 𝑧 ∈ ( V ∖ { 𝑧 ∣ 𝜑 } ) ↔ 𝑧 ∈ { 𝑧 ∣ ¬ 𝜑 } ) | 
| 11 | 5 10 | mpgbir | ⊢ ( V ∖ { 𝑧 ∣ 𝜑 } ) = { 𝑧 ∣ ¬ 𝜑 } |