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Mirrors > Home > MPE Home > Th. List > sbss | Structured version Visualization version GIF version |
Description: Set substitution into the first argument of a subset relation. (Contributed by Rodolfo Medina, 7-Jul-2010.) (Proof shortened by Mario Carneiro, 14-Nov-2016.) |
Ref | Expression |
---|---|
sbss | ⊢ ([𝑦 / 𝑥]𝑥 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | sseq1 4021 | . 2 ⊢ (𝑥 = 𝑧 → (𝑥 ⊆ 𝐴 ↔ 𝑧 ⊆ 𝐴)) | |
2 | sseq1 4021 | . 2 ⊢ (𝑧 = 𝑦 → (𝑧 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴)) | |
3 | 1, 2 | sbievw2 2096 | 1 ⊢ ([𝑦 / 𝑥]𝑥 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 206 [wsb 2062 ⊆ wss 3963 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1792 ax-4 1806 ax-5 1908 ax-6 1965 ax-7 2005 ax-9 2116 ax-ext 2706 |
This theorem depends on definitions: df-bi 207 df-an 396 df-ex 1777 df-sb 2063 df-cleq 2727 df-ss 3980 |
This theorem is referenced by: (None) |
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