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Theorem relss 4686
Description: Subclass theorem for relation predicate. Theorem 2 of [Suppes] p. 58. (Contributed by NM, 15-Aug-1994.)
Assertion
Ref Expression
relss (𝐴𝐵 → (Rel 𝐵 → Rel 𝐴))

Proof of Theorem relss
StepHypRef Expression
1 sstr2 3145 . 2 (𝐴𝐵 → (𝐵 ⊆ (V × V) → 𝐴 ⊆ (V × V)))
2 df-rel 4606 . 2 (Rel 𝐵𝐵 ⊆ (V × V))
3 df-rel 4606 . 2 (Rel 𝐴𝐴 ⊆ (V × V))
41, 2, 33imtr4g 204 1 (𝐴𝐵 → (Rel 𝐵 → Rel 𝐴))
Colors of variables: wff set class
Syntax hints:  wi 4  Vcvv 2722  wss 3112   × cxp 4597  Rel wrel 4604
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1434  ax-7 1435  ax-gen 1436  ax-ie1 1480  ax-ie2 1481  ax-8 1491  ax-11 1493  ax-4 1497  ax-17 1513  ax-i9 1517  ax-ial 1521  ax-i5r 1522  ax-ext 2146
This theorem depends on definitions:  df-bi 116  df-nf 1448  df-sb 1750  df-clab 2151  df-cleq 2157  df-clel 2160  df-in 3118  df-ss 3125  df-rel 4606
This theorem is referenced by:  relin1  4717  relin2  4718  reldif  4719  relres  4907  iss  4925  cnvdif  5005  funss  5202  funssres  5225  fliftcnv  5758  fliftfun  5759  reltpos  6210  tpostpos  6224  swoer  6521  erinxp  6567  ltrel  7952  lerel  7954  txdis1cn  12845  xmeter  13003
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