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Theorem sbceq1g 2864
Description: Move proper substitution to first argument of an equality. (Contributed by NM, 30-Nov-2005.)
Assertion
Ref Expression
sbceq1g (A 𝑉 → ([A / x]B = 𝐶A / xB = 𝐶))
Distinct variable group:   x,𝐶
Allowed substitution hints:   A(x)   B(x)   𝑉(x)

Proof of Theorem sbceq1g
StepHypRef Expression
1 sbceqg 2860 . 2 (A 𝑉 → ([A / x]B = 𝐶A / xB = A / x𝐶))
2 csbconstg 2858 . . 3 (A 𝑉A / x𝐶 = 𝐶)
32eqeq2d 2048 . 2 (A 𝑉 → (A / xB = A / x𝐶A / xB = 𝐶))
41, 3bitrd 177 1 (A 𝑉 → ([A / x]B = 𝐶A / xB = 𝐶))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 98   = wceq 1242   wcel 1390  [wsbc 2758  csb 2846
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-sbc 2759  df-csb 2847
This theorem is referenced by: (None)
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