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Theorem dfixp 8482
 Description: Eliminate the expression {𝑥 ∣ 𝑥 ∈ 𝐴} in df-ixp 8481, under the assumption that 𝐴 and 𝑥 are disjoint. This way, we can say that 𝑥 is bound in X𝑥 ∈ 𝐴𝐵 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 8481 . 2 X𝑥𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
2 abid2 2895 . . . . 5 {𝑥𝑥𝐴} = 𝐴
32fneq2i 6433 . . . 4 (𝑓 Fn {𝑥𝑥𝐴} ↔ 𝑓 Fn 𝐴)
43anbi1i 627 . . 3 ((𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵) ↔ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵))
54abbii 2824 . 2 {𝑓 ∣ (𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)} = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
61, 5eqtri 2782 1 X𝑥𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
 Colors of variables: wff setvar class Syntax hints:   ∧ wa 400   = wceq 1539   ∈ wcel 2112  {cab 2736  ∀wral 3071   Fn wfn 6331  ‘cfv 6336  Xcixp 8480 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 1912  ax-6 1971  ax-7 2016  ax-8 2114  ax-9 2122  ax-ext 2730 This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1542  df-ex 1783  df-sb 2071  df-clab 2737  df-cleq 2751  df-clel 2831  df-fn 6339  df-ixp 8481 This theorem is referenced by:  ixpsnval  8483  elixp2  8484  ixpeq1  8491  cbvixp  8497  ixp0x  8509
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