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Theorem dfixp 8897
Description: Eliminate the expression {𝑥𝑥𝐴} in df-ixp 8896, 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 8896 . 2 X𝑥𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
2 abid2 2906 . . . . 5 {𝑥𝑥𝐴} = 𝐴
32fneq2i 6634 . . . 4 (𝑓 Fn {𝑥𝑥𝐴} ↔ 𝑓 Fn 𝐴)
43anbi1i 635 . . 3 ((𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵) ↔ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵))
54abbii 2836 . 2 {𝑓 ∣ (𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)} = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
61, 5eqtri 2792 1 X𝑥𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
Colors of variables: wff setvar class
Syntax hints:  wa 400   = wceq 1567  wcel 2149  {cab 2747  wral 3085   Fn wfn 6532  cfv 6537  Xcixp 8895
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741
This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1570  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-fn 6540  df-ixp 8896
This theorem is referenced by:  ixpsnval  8898  elixp2  8899  ixpeq1  8906  cbvixp  8912  cbvixpv  8913  ixp0x  8924
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