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Theorem ffdm 6699
Description: A mapping is a partial function. (Contributed by NM, 25-Nov-2007.)
Assertion
Ref Expression
ffdm (𝐹:𝐴𝐵 → (𝐹:dom 𝐹𝐵 ∧ dom 𝐹𝐴))

Proof of Theorem ffdm
StepHypRef Expression
1 fdm 6678 . . . 4 (𝐹:𝐴𝐵 → dom 𝐹 = 𝐴)
21feq2d 6655 . . 3 (𝐹:𝐴𝐵 → (𝐹:dom 𝐹𝐵𝐹:𝐴𝐵))
32ibir 268 . 2 (𝐹:𝐴𝐵𝐹:dom 𝐹𝐵)
4 eqimss 4001 . . 3 (dom 𝐹 = 𝐴 → dom 𝐹𝐴)
51, 4syl 17 . 2 (𝐹:𝐴𝐵 → dom 𝐹𝐴)
63, 5jca 513 1 (𝐹:𝐴𝐵 → (𝐹:dom 𝐹𝐵 ∧ dom 𝐹𝐴))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 397   = wceq 1542  wss 3911  dom cdm 5634  wf 6493
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704
This theorem depends on definitions:  df-bi 206  df-an 398  df-tru 1545  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-v 3446  df-in 3918  df-ss 3928  df-fn 6500  df-f 6501
This theorem is referenced by:  ffdmd  6700  smoiso  8309  s4f1o  14813  islindf2  21236  fourierdlem92  44525  fouriersw  44558  etransclem2  44563
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