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Mirrors > Home > MPE Home > Th. List > euequ1 | Structured version Visualization version GIF version |
Description: Equality has existential uniqueness. Special case of eueq1 3520 proved using only predicate calculus. The proof needs 𝑦 = 𝑧 be free of 𝑥. This is ensured by having 𝑥 and 𝑦 be distinct. Alternately, a distinctor ¬ ∀𝑥𝑥 = 𝑦 could have been used instead. (Contributed by Stefan Allan, 4-Dec-2008.) (Proof shortened by Wolf Lammen, 8-Sep-2019.) |
Ref | Expression |
---|---|
euequ1 | ⊢ ∃!𝑥 𝑥 = 𝑦 |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | ax6evr 2097 | . . 3 ⊢ ∃𝑧 𝑦 = 𝑧 | |
2 | equequ2 2108 | . . . 4 ⊢ (𝑦 = 𝑧 → (𝑥 = 𝑦 ↔ 𝑥 = 𝑧)) | |
3 | 2 | alrimiv 2004 | . . 3 ⊢ (𝑦 = 𝑧 → ∀𝑥(𝑥 = 𝑦 ↔ 𝑥 = 𝑧)) |
4 | 1, 3 | eximii 1913 | . 2 ⊢ ∃𝑧∀𝑥(𝑥 = 𝑦 ↔ 𝑥 = 𝑧) |
5 | df-eu 2611 | . 2 ⊢ (∃!𝑥 𝑥 = 𝑦 ↔ ∃𝑧∀𝑥(𝑥 = 𝑦 ↔ 𝑥 = 𝑧)) | |
6 | 4, 5 | mpbir 221 | 1 ⊢ ∃!𝑥 𝑥 = 𝑦 |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 196 ∀wal 1630 ∃wex 1853 ∃!weu 2607 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1871 ax-4 1886 ax-5 1988 ax-6 2054 ax-7 2090 |
This theorem depends on definitions: df-bi 197 df-an 385 df-ex 1854 df-eu 2611 |
This theorem is referenced by: copsexg 5104 oprabid 6840 |
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