| Mathbox for Richard Penner |
< Previous
Next >
Nearby theorems |
||
| 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 4646 | . 2 ⊢ {𝑌, 𝑍} = {𝑥 ∣ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍)} | |
| 2 | 1 | abeqabi 43421 | 1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌, 𝑍} ↔ ∀𝑥(𝜑 ↔ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍))) |
| Colors of variables: wff setvar class |
| Syntax hints: ↔ wb 206 ∨ wo 848 ∀wal 1538 = wceq 1540 {cab 2714 {cpr 4628 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1795 ax-4 1809 ax-5 1910 ax-6 1967 ax-7 2007 ax-8 2110 ax-9 2118 ax-10 2141 ax-11 2157 ax-12 2177 ax-ext 2708 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 849 df-tru 1543 df-ex 1780 df-nf 1784 df-sb 2065 df-clab 2715 df-cleq 2729 df-clel 2816 df-v 3482 df-un 3956 df-sn 4627 df-pr 4629 |
| This theorem is referenced by: (None) |
| Copyright terms: Public domain | W3C validator |