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Theorem ineqri 3127
Description: Inference from membership to intersection. (Contributed by NM, 5-Aug-1993.)
Hypothesis
Ref Expression
ineqri.1 ((𝑥𝐴𝑥𝐵) ↔ 𝑥𝐶)
Assertion
Ref Expression
ineqri (𝐴𝐵) = 𝐶
Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝑥,𝐶

Proof of Theorem ineqri
StepHypRef Expression
1 elin 3123 . . 3 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
2 ineqri.1 . . 3 ((𝑥𝐴𝑥𝐵) ↔ 𝑥𝐶)
31, 2bitri 173 . 2 (𝑥 ∈ (𝐴𝐵) ↔ 𝑥𝐶)
43eqriv 2037 1 (𝐴𝐵) = 𝐶
Colors of variables: wff set class
Syntax hints:  wa 97  wb 98   = wceq 1243  wcel 1393  cin 2913
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 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022
This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2556  df-in 2921
This theorem is referenced by:  inidm  3143  inass  3144  indi  3181  inab  3202  in0  3249  pwin  4016  dmres  4619
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