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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 4576 | . 2 ⊢ {𝑌, 𝑍} = {𝑥 ∣ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍)} | |
| 2 | 1 | abeqabi 43852 | 1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌, 𝑍} ↔ ∀𝑥(𝜑 ↔ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍))) |
| Colors of variables: wff setvar class |
| Syntax hints: ↔ wb 207 ∨ wo 853 ∀wal 1545 = wceq 1547 {cab 2717 {cpr 4557 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1802 ax-4 1816 ax-5 1917 ax-6 1974 ax-7 2015 ax-8 2121 ax-9 2129 ax-10 2152 ax-11 2168 ax-12 2189 ax-ext 2711 |
| This theorem depends on definitions: df-bi 208 df-an 397 df-or 854 df-tru 1550 df-ex 1787 df-nf 1791 df-sb 2074 df-clab 2718 df-cleq 2731 df-clel 2814 df-v 3433 df-un 3888 df-sn 4556 df-pr 4558 |
| This theorem is referenced by: (None) |
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