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| Mirrors > Home > MPE Home > Th. List > Mathboxes > sbeqal1i | Structured version Visualization version GIF version | ||
| Description: Suppose you know 𝑥 = 𝑦 implies 𝑥 = 𝑧, assuming 𝑥 and 𝑧 are distinct. Then, 𝑦 = 𝑧. (Contributed by Andrew Salmon, 3-Jun-2011.) | 
| Ref | Expression | 
|---|---|
| sbeqal1i.1 | ⊢ (𝑥 = 𝑦 → 𝑥 = 𝑧) | 
| Ref | Expression | 
|---|---|
| sbeqal1i | ⊢ 𝑦 = 𝑧 | 
| Step | Hyp | Ref | Expression | 
|---|---|---|---|
| 1 | sbeqal1 44417 | . 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝑥 = 𝑧) → 𝑦 = 𝑧) | |
| 2 | sbeqal1i.1 | . 2 ⊢ (𝑥 = 𝑦 → 𝑥 = 𝑧) | |
| 3 | 1, 2 | mpg 1797 | 1 ⊢ 𝑦 = 𝑧 | 
| Colors of variables: wff setvar class | 
| Syntax hints: → wi 4 | 
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1795 ax-4 1809 ax-5 1910 ax-6 1967 ax-7 2007 ax-10 2141 ax-12 2177 ax-13 2377 | 
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 849 df-ex 1780 df-nf 1784 df-sb 2065 | 
| This theorem is referenced by: sbeqal2i 44419 | 
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