Theorem dfixp 8938

Description: Eliminate the expression {𝑥 ∣ 𝑥 ∈ 𝐴} in df-ixp 8937, 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	⊢ X𝑥 ∈ 𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)}

Distinct variable groups:   𝑥,𝑓,𝐴   𝐵,𝑓   𝑥,𝐴

Allowed substitution hint:   𝐵(𝑥)

Proof of Theorem dfixp

Step	Hyp	Ref	Expression
1		df-ixp 8937	. 2 ⊢ X𝑥 ∈ 𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)}
2		abid2 2877	. . . . 5 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
3	2	fneq2i 6667	. . . 4 ⊢ (𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ 𝑓 Fn 𝐴)
4	3	anbi1i 624	. . 3 ⊢ ((𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵) ↔ (𝑓 Fn 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵))
5	4	abbii 2807	. 2 ⊢ {𝑓 ∣ (𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)} = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)}
6	1, 5	eqtri 2763	1 ⊢ X𝑥 ∈ 𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)}

Syntax hints:   ∧ wa 395   = wceq 1537   ∈ wcel 2106  {cab 2712  ∀wral 3059   Fn wfn 6558  ‘cfv 6563  Xcixp 8936

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-fn 6566  df-ixp 8937

This theorem is referenced by:  ixpsnval  8939  elixp2  8940  ixpeq1  8947  cbvixp  8953  cbvixpv  8954  ixp0x  8965