![]() |
Metamath Proof Explorer |
< Previous
Next >
Nearby theorems |
|
Mirrors > Home > MPE Home > Th. List > absn | Structured version Visualization version GIF version |
Description: Condition for a class abstraction to be a singleton. Formerly part of proof of dfiota2 6065. (Contributed by Andrew Salmon, 30-Jun-2011.) (Revised by AV, 24-Aug-2022.) |
Ref | Expression |
---|---|
absn | ⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑌)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-sn 4369 | . . 3 ⊢ {𝑌} = {𝑥 ∣ 𝑥 = 𝑌} | |
2 | 1 | eqeq2i 2811 | . 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝑥 = 𝑌}) |
3 | abbi 2914 | . 2 ⊢ (∀𝑥(𝜑 ↔ 𝑥 = 𝑌) ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝑥 = 𝑌}) | |
4 | 2, 3 | bitr4i 270 | 1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑌)) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 198 ∀wal 1651 = wceq 1653 {cab 2785 {csn 4368 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1891 ax-4 1905 ax-5 2006 ax-6 2072 ax-7 2107 ax-9 2166 ax-10 2185 ax-11 2200 ax-12 2213 ax-ext 2777 |
This theorem depends on definitions: df-bi 199 df-an 386 df-or 875 df-tru 1657 df-ex 1876 df-nf 1880 df-sb 2065 df-clab 2786 df-cleq 2792 df-clel 2795 df-sn 4369 |
This theorem is referenced by: rabeqsn 4405 dfiota2 6065 dfaiota2 41935 aiotaval 41942 |
Copyright terms: Public domain | W3C validator |