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Theorem sbcan 2799
Description: Distribution of class substitution over conjunction. (Contributed by NM, 31-Dec-2016.)
Assertion
Ref Expression
sbcan ([A / x](φ ψ) ↔ ([A / x]φ [A / x]ψ))

Proof of Theorem sbcan
Dummy variable y is distinct from all other variables.
StepHypRef Expression
1 sbcex 2766 . 2 ([A / x](φ ψ) → A V)
2 sbcex 2766 . . 3 ([A / x]ψA V)
32adantl 262 . 2 (([A / x]φ [A / x]ψ) → A V)
4 dfsbcq2 2761 . . 3 (y = A → ([y / x](φ ψ) ↔ [A / x](φ ψ)))
5 dfsbcq2 2761 . . . 4 (y = A → ([y / x]φ[A / x]φ))
6 dfsbcq2 2761 . . . 4 (y = A → ([y / x]ψ[A / x]ψ))
75, 6anbi12d 442 . . 3 (y = A → (([y / x]φ [y / x]ψ) ↔ ([A / x]φ [A / x]ψ)))
8 sban 1826 . . 3 ([y / x](φ ψ) ↔ ([y / x]φ [y / x]ψ))
94, 7, 8vtoclbg 2608 . 2 (A V → ([A / x](φ ψ) ↔ ([A / x]φ [A / x]ψ)))
101, 3, 9pm5.21nii 619 1 ([A / x](φ ψ) ↔ ([A / x]φ [A / x]ψ))
Colors of variables: wff set class
Syntax hints:   wa 97  wb 98   = wceq 1242   wcel 1390  [wsb 1642  Vcvv 2551  [wsbc 2758
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
This theorem is referenced by:  difopab  4412  sbcfung  4868  sbcfng  4987  sbcfg  4988
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