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Theorem sbcrel 5789
Description: Distribute proper substitution through a relation predicate. (Contributed by Alexander van der Vekens, 23-Jul-2017.)
Assertion
Ref Expression
sbcrel (𝐴𝑉 → ([𝐴 / 𝑥]Rel 𝑅 ↔ Rel 𝐴 / 𝑥𝑅))

Proof of Theorem sbcrel
StepHypRef Expression
1 sbcssg 4519 . . 3 (𝐴𝑉 → ([𝐴 / 𝑥]𝑅 ⊆ (V × V) ↔ 𝐴 / 𝑥𝑅𝐴 / 𝑥(V × V)))
2 csbconstg 3917 . . . 4 (𝐴𝑉𝐴 / 𝑥(V × V) = (V × V))
32sseq2d 4015 . . 3 (𝐴𝑉 → (𝐴 / 𝑥𝑅𝐴 / 𝑥(V × V) ↔ 𝐴 / 𝑥𝑅 ⊆ (V × V)))
41, 3bitrd 279 . 2 (𝐴𝑉 → ([𝐴 / 𝑥]𝑅 ⊆ (V × V) ↔ 𝐴 / 𝑥𝑅 ⊆ (V × V)))
5 df-rel 5691 . . 3 (Rel 𝑅𝑅 ⊆ (V × V))
65sbcbii 3845 . 2 ([𝐴 / 𝑥]Rel 𝑅[𝐴 / 𝑥]𝑅 ⊆ (V × V))
7 df-rel 5691 . 2 (Rel 𝐴 / 𝑥𝑅𝐴 / 𝑥𝑅 ⊆ (V × V))
84, 6, 73bitr4g 314 1 (𝐴𝑉 → ([𝐴 / 𝑥]Rel 𝑅 ↔ Rel 𝐴 / 𝑥𝑅))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wcel 2107  Vcvv 3479  [wsbc 3787  csb 3898  wss 3950   × cxp 5682  Rel wrel 5689
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-10 2140  ax-11 2156  ax-12 2176  ax-ext 2707
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1542  df-fal 1552  df-ex 1779  df-nf 1783  df-sb 2064  df-clab 2714  df-cleq 2728  df-clel 2815  df-nfc 2891  df-v 3481  df-sbc 3788  df-csb 3899  df-dif 3953  df-ss 3967  df-nul 4333  df-rel 5691
This theorem is referenced by:  sbcfung  6589
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