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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 4668 | . 2 ⊢ {𝑌, 𝑍} = {𝑥 ∣ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍)} | |
2 | 1 | abeqabi 43365 | 1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌, 𝑍} ↔ ∀𝑥(𝜑 ↔ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍))) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 206 ∨ wo 846 ∀wal 1535 = wceq 1537 {cab 2717 {cpr 4650 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1793 ax-4 1807 ax-5 1909 ax-6 1967 ax-7 2007 ax-8 2110 ax-9 2118 ax-10 2141 ax-11 2158 ax-12 2178 ax-ext 2711 |
This theorem depends on definitions: df-bi 207 df-an 396 df-or 847 df-tru 1540 df-ex 1778 df-nf 1782 df-sb 2065 df-clab 2718 df-cleq 2732 df-clel 2819 df-v 3490 df-un 3981 df-sn 4649 df-pr 4651 |
This theorem is referenced by: (None) |
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