![]() |
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 4644 | . 2 ⊢ {𝑌, 𝑍} = {𝑥 ∣ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍)} | |
2 | 1 | abeqabi 42829 | 1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌, 𝑍} ↔ ∀𝑥(𝜑 ↔ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍))) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 205 ∨ wo 846 ∀wal 1532 = wceq 1534 {cab 2705 {cpr 4627 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1790 ax-4 1804 ax-5 1906 ax-6 1964 ax-7 2004 ax-8 2101 ax-9 2109 ax-10 2130 ax-11 2147 ax-12 2167 ax-ext 2699 |
This theorem depends on definitions: df-bi 206 df-an 396 df-or 847 df-tru 1537 df-ex 1775 df-nf 1779 df-sb 2061 df-clab 2706 df-cleq 2720 df-clel 2806 df-v 3472 df-un 3950 df-sn 4626 df-pr 4628 |
This theorem is referenced by: (None) |
Copyright terms: Public domain | W3C validator |