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| Mirrors > Home > MPE Home > Th. List > Mathboxes > abtp | Structured version Visualization version GIF version | ||
| Description: Condition for a class abstraction to be a triple. (Contributed by RP, 25-Aug-2024.) |
| Ref | Expression |
|---|---|
| abtp | ⊢ ({𝑥 ∣ 𝜑} = {𝑋, 𝑌, 𝑍} ↔ ∀𝑥(𝜑 ↔ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍))) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | dftp2 4641 | . 2 ⊢ {𝑋, 𝑌, 𝑍} = {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)} | |
| 2 | 1 | abeqabi 43449 | 1 ⊢ ({𝑥 ∣ 𝜑} = {𝑋, 𝑌, 𝑍} ↔ ∀𝑥(𝜑 ↔ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍))) |
| Colors of variables: wff setvar class |
| Syntax hints: ↔ wb 206 ∨ w3o 1085 ∀wal 1539 = wceq 1541 {cab 2709 {ctp 4577 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1968 ax-7 2009 ax-8 2113 ax-9 2121 ax-10 2144 ax-11 2160 ax-12 2180 ax-ext 2703 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-3or 1087 df-tru 1544 df-ex 1781 df-nf 1785 df-sb 2068 df-clab 2710 df-cleq 2723 df-clel 2806 df-v 3438 df-un 3902 df-sn 4574 df-pr 4576 df-tp 4578 |
| This theorem is referenced by: (None) |
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