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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 4651 | . 2 ⊢ {𝑋, 𝑌, 𝑍} = {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)} | |
| 2 | 1 | abeqabi 44549 | 1 ⊢ ({𝑥 ∣ 𝜑} = {𝑋, 𝑌, 𝑍} ↔ ∀𝑥(𝜑 ↔ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍))) |
| Colors of variables: wff setvar class |
| This proof depends on syntax axioms: ↔ wb 209 ∨ w3o 1102 ∀wal 1567 = wceq 1569 {cab 2741 {ctp 4591 |
| This proof depends on axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1827 ax-4 1841 ax-5 1944 ax-6 2001 ax-7 2042 ax-8 2148 ax-9 2156 ax-10 2179 ax-11 2196 ax-12 2215 ax-ext 2735 |
| This proof depends on definitions: df-bi 210 df-an 402 df-or 862 df-3or 1104 df-tru 1572 df-ex 1812 df-nf 1816 df-sb 2101 df-clab 2742 df-cleq 2755 df-clel 2838 df-v 3455 df-un 4153 df-sn 4588 df-pr 4590 df-tp 4592 |
| This theorem is used by: (None) |
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