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Mathbox for Andrew Salmon |
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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 42770 | . 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝑥 = 𝑧) → 𝑦 = 𝑧) | |
2 | sbeqal1i.1 | . 2 ⊢ (𝑥 = 𝑦 → 𝑥 = 𝑧) | |
3 | 1, 2 | mpg 1800 | 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 1798 ax-4 1812 ax-5 1914 ax-6 1972 ax-7 2012 ax-10 2138 ax-12 2172 ax-13 2371 |
This theorem depends on definitions: df-bi 206 df-an 398 df-or 847 df-ex 1783 df-nf 1787 df-sb 2069 |
This theorem is referenced by: sbeqal2i 42772 |
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