Theorem dfixp 6912

Description: Eliminate the expression {𝑥 ∣ 𝑥 ∈ 𝐴} in df-ixp 6911, 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 6911	. 2 ⊢ X𝑥 ∈ 𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)}
2		abid2 2353	. . . . 5 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
3	2	fneq2i 5432	. . . 4 ⊢ (𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ 𝑓 Fn 𝐴)
4	3	anbi1i 458	. . 3 ⊢ ((𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵) ↔ (𝑓 Fn 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵))
5	4	abbii 2347	. 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 2252	1 ⊢ X𝑥 ∈ 𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)}

Syntax hints:   ∧ wa 104   = wceq 1398   ∈ wcel 2202  {cab 2217  ∀wral 2511   Fn wfn 5328  ‘cfv 5333  Xcixp 6910

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-11 1555  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-fn 5336  df-ixp 6911

This theorem is referenced by:  ixpsnval  6913  elixp2  6914  ixpeq1  6921  cbvixp  6927  ixp0x  6938