Theorem dfixp 8899

Description: Eliminate the expression {𝑥 ∣ 𝑥 ∈ 𝐴} in df-ixp 8898, 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 8898	. 2 ⊢ X𝑥 ∈ 𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)}
2		abid2 2870	. . . . 5 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
3	2	fneq2i 6647	. . . 4 ⊢ (𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ 𝑓 Fn 𝐴)
4	3	anbi1i 623	. . 3 ⊢ ((𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵) ↔ (𝑓 Fn 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵))
5	4	abbii 2801	. 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 2759	1 ⊢ X𝑥 ∈ 𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)}

Syntax hints:   ∧ wa 395   = wceq 1540   ∈ wcel 2105  {cab 2708  ∀wral 3060   Fn wfn 6538  ‘cfv 6543  Xcixp 8897

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1543  df-ex 1781  df-sb 2067  df-clab 2709  df-cleq 2723  df-clel 2809  df-fn 6546  df-ixp 8898

This theorem is referenced by:  ixpsnval  8900  elixp2  8901  ixpeq1  8908  cbvixp  8914  ixp0x  8926