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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 4610 | . 2 ⊢ {𝑌, 𝑍} = {𝑥 ∣ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍)} | |
2 | 1 | abeqabi 41754 | 1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌, 𝑍} ↔ ∀𝑥(𝜑 ↔ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍))) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 205 ∨ wo 846 ∀wal 1540 = wceq 1542 {cab 2714 {cpr 4593 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1798 ax-4 1812 ax-5 1914 ax-6 1972 ax-7 2012 ax-8 2109 ax-9 2117 ax-10 2138 ax-11 2155 ax-12 2172 ax-ext 2708 |
This theorem depends on definitions: df-bi 206 df-an 398 df-or 847 df-tru 1545 df-ex 1783 df-nf 1787 df-sb 2069 df-clab 2715 df-cleq 2729 df-clel 2815 df-v 3450 df-un 3920 df-sn 4592 df-pr 4594 |
This theorem is referenced by: (None) |
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