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Theorem funeqd 4951
Description: Equality deduction for the function predicate. (Contributed by NM, 23-Feb-2013.)
Hypothesis
Ref Expression
funeqd.1 (𝜑𝐴 = 𝐵)
Assertion
Ref Expression
funeqd (𝜑 → (Fun 𝐴 ↔ Fun 𝐵))

Proof of Theorem funeqd
StepHypRef Expression
1 funeqd.1 . 2 (𝜑𝐴 = 𝐵)
2 funeq 4949 . 2 (𝐴 = 𝐵 → (Fun 𝐴 ↔ Fun 𝐵))
31, 2syl 14 1 (𝜑 → (Fun 𝐴 ↔ Fun 𝐵))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 102   = wceq 1259  Fun wfun 4924
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038
This theorem depends on definitions:  df-bi 114  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-nfc 2183  df-in 2952  df-ss 2959  df-br 3793  df-opab 3847  df-rel 4380  df-cnv 4381  df-co 4382  df-fun 4932
This theorem is referenced by:  funopg  4962  funsng  4974  funcnvuni  4996  f1eq1  5115  shftfn  9653
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