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Theorem fndmu 5000
Description: A function has a unique domain. (Contributed by NM, 11-Aug-1994.)
Assertion
Ref Expression
fndmu ((𝐹 Fn 𝐴𝐹 Fn 𝐵) → 𝐴 = 𝐵)

Proof of Theorem fndmu
StepHypRef Expression
1 fndm 4998 . 2 (𝐹 Fn 𝐴 → dom 𝐹 = 𝐴)
2 fndm 4998 . 2 (𝐹 Fn 𝐵 → dom 𝐹 = 𝐵)
31, 2sylan9req 2093 1 ((𝐹 Fn 𝐴𝐹 Fn 𝐵) → 𝐴 = 𝐵)
Colors of variables: wff set class
Syntax hints:  wi 4  wa 97   = wceq 1243  dom cdm 4345   Fn wfn 4897
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-5 1336  ax-gen 1338  ax-4 1400  ax-17 1419  ax-ext 2022
This theorem depends on definitions:  df-bi 110  df-cleq 2033  df-fn 4905
This theorem is referenced by:  fodmrnu  5114  tfrlemisucaccv  5939  0fz1  8909
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