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Theorem dfixp 6713
Description: Eliminate the expression {𝑥𝑥𝐴} in df-ixp 6712, under the assumption that 𝐴 and 𝑥 are disjoint. This way, we can say that 𝑥 is bound in X𝑥𝐴𝐵 even if it appears free in 𝐴. (Contributed by Mario Carneiro, 12-Aug-2016.)
Assertion
Ref Expression
dfixp X𝑥𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
Distinct variable groups:   𝑥,𝑓,𝐴   𝐵,𝑓   𝑥,𝐴
Allowed substitution hint:   𝐵(𝑥)

Proof of Theorem dfixp
StepHypRef Expression
1 df-ixp 6712 . 2 X𝑥𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
2 abid2 2308 . . . . 5 {𝑥𝑥𝐴} = 𝐴
32fneq2i 5323 . . . 4 (𝑓 Fn {𝑥𝑥𝐴} ↔ 𝑓 Fn 𝐴)
43anbi1i 458 . . 3 ((𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵) ↔ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵))
54abbii 2303 . 2 {𝑓 ∣ (𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)} = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
61, 5eqtri 2208 1 X𝑥𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
Colors of variables: wff set class
Syntax hints:  wa 104   = wceq 1363  wcel 2158  {cab 2173  wral 2465   Fn wfn 5223  cfv 5228  Xcixp 6711
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1457  ax-7 1458  ax-gen 1459  ax-ie1 1503  ax-ie2 1504  ax-8 1514  ax-11 1516  ax-4 1520  ax-17 1536  ax-i9 1540  ax-ial 1544  ax-i5r 1545  ax-ext 2169
This theorem depends on definitions:  df-bi 117  df-tru 1366  df-nf 1471  df-sb 1773  df-clab 2174  df-cleq 2180  df-clel 2183  df-fn 5231  df-ixp 6712
This theorem is referenced by:  ixpsnval  6714  elixp2  6715  ixpeq1  6722  cbvixp  6728  ixp0x  6739
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