Intuitionistic Logic Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  ILE Home  >  Th. List  >  dfixp GIF version

Theorem dfixp 6497
 Description: Eliminate the expression {𝑥 ∣ 𝑥 ∈ 𝐴} in df-ixp 6496, 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 6496 . 2 X𝑥𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
2 abid2 2215 . . . . 5 {𝑥𝑥𝐴} = 𝐴
32fneq2i 5143 . . . 4 (𝑓 Fn {𝑥𝑥𝐴} ↔ 𝑓 Fn 𝐴)
43anbi1i 447 . . 3 ((𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵) ↔ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵))
54abbii 2210 . 2 {𝑓 ∣ (𝑓 Fn {𝑥𝑥𝐴} ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)} = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
61, 5eqtri 2115 1 X𝑥𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥𝐴 (𝑓𝑥) ∈ 𝐵)}
 Colors of variables: wff set class Syntax hints:   ∧ wa 103   = wceq 1296   ∈ wcel 1445  {cab 2081  ∀wral 2370   Fn wfn 5044  ‘cfv 5049  Xcixp 6495 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-11 1449  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077 This theorem depends on definitions:  df-bi 116  df-tru 1299  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-fn 5052  df-ixp 6496 This theorem is referenced by:  ixpsnval  6498  elixp2  6499  ixpeq1  6506  cbvixp  6512  ixp0x  6523
 Copyright terms: Public domain W3C validator