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Theorem ssrd 3197
Description: Deduction based on subclass definition. (Contributed by Thierry Arnoux, 8-Mar-2017.)
Hypotheses
Ref Expression
ssrd.0 𝑥𝜑
ssrd.1 𝑥𝐴
ssrd.2 𝑥𝐵
ssrd.3 (𝜑 → (𝑥𝐴𝑥𝐵))
Assertion
Ref Expression
ssrd (𝜑𝐴𝐵)

Proof of Theorem ssrd
StepHypRef Expression
1 ssrd.0 . . 3 𝑥𝜑
2 ssrd.3 . . 3 (𝜑 → (𝑥𝐴𝑥𝐵))
31, 2alrimi 1544 . 2 (𝜑 → ∀𝑥(𝑥𝐴𝑥𝐵))
4 ssrd.1 . . 3 𝑥𝐴
5 ssrd.2 . . 3 𝑥𝐵
64, 5dfss2f 3183 . 2 (𝐴𝐵 ↔ ∀𝑥(𝑥𝐴𝑥𝐵))
73, 6sylibr 134 1 (𝜑𝐴𝐵)
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1370  wnf 1482  wcel 2175  wnfc 2334  wss 3165
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1469  ax-7 1470  ax-gen 1471  ax-ie1 1515  ax-ie2 1516  ax-8 1526  ax-10 1527  ax-11 1528  ax-i12 1529  ax-bndl 1531  ax-4 1532  ax-17 1548  ax-i9 1552  ax-ial 1556  ax-i5r 1557  ax-ext 2186
This theorem depends on definitions:  df-bi 117  df-nf 1483  df-sb 1785  df-clab 2191  df-cleq 2197  df-clel 2200  df-nfc 2336  df-in 3171  df-ss 3178
This theorem is referenced by:  eqrd  3210  exmidomni  7226
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