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Theorem dfixp 8840
Description: Eliminate the expression {𝑥𝑥𝐴} in df-ixp 8839, 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 8839 . 2 X𝑥𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
2 abid2 2872 . . . . 5 {𝑥𝑥𝐴} = 𝐴
32fneq2i 6601 . . . 4 (𝑓 Fn {𝑥𝑥𝐴} ↔ 𝑓 Fn 𝐴)
43anbi1i 625 . . 3 ((𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵) ↔ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵))
54abbii 2803 . 2 {𝑓 ∣ (𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)} = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
61, 5eqtri 2761 1 X𝑥𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
Colors of variables: wff setvar class
Syntax hints:  wa 397   = wceq 1542  wcel 2107  {cab 2710  wral 3061   Fn wfn 6492  cfv 6497  Xcixp 8838
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-fn 6500  df-ixp 8839
This theorem is referenced by:  ixpsnval  8841  elixp2  8842  ixpeq1  8849  cbvixp  8855  ixp0x  8867
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