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| Mirrors > Home > MPE Home > Th. List > Mathboxes > abpr | Structured version Visualization version GIF version | ||
| Description: Condition for a class abstraction to be a pair. (Contributed by RP, 25-Aug-2024.) |
| Ref | Expression |
|---|---|
| abpr | ⊢ ({𝑥 ∣ 𝜑} = {𝑌, 𝑍} ↔ ∀𝑥(𝜑 ↔ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍))) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | dfpr2 4603 | . 2 ⊢ {𝑌, 𝑍} = {𝑥 ∣ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍)} | |
| 2 | 1 | abeqabi 43764 | 1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌, 𝑍} ↔ ∀𝑥(𝜑 ↔ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍))) |
| Colors of variables: wff setvar class |
| Syntax hints: ↔ wb 206 ∨ wo 848 ∀wal 1540 = wceq 1542 {cab 2715 {cpr 4584 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1912 ax-6 1969 ax-7 2010 ax-8 2116 ax-9 2124 ax-10 2147 ax-11 2163 ax-12 2185 ax-ext 2709 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 849 df-tru 1545 df-ex 1782 df-nf 1786 df-sb 2069 df-clab 2716 df-cleq 2729 df-clel 2812 df-v 3444 df-un 3908 df-sn 4583 df-pr 4585 |
| This theorem is referenced by: (None) |
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