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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 3940 | . 2 ⊢ (𝑥 = 𝑧 → (𝑥 ⊆ 𝐴 ↔ 𝑧 ⊆ 𝐴)) | |
2 | sseq1 3940 | . 2 ⊢ (𝑧 = 𝑦 → (𝑧 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴)) | |
3 | 1, 2 | sbievw2 2104 | 1 ⊢ ([𝑦 / 𝑥]𝑥 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 209 [wsb 2069 ⊆ wss 3881 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2113 ax-9 2121 ax-ext 2770 |
This theorem depends on definitions: df-bi 210 df-an 400 df-ex 1782 df-sb 2070 df-clab 2777 df-cleq 2791 df-clel 2870 df-v 3443 df-in 3888 df-ss 3898 |
This theorem is referenced by: (None) |
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