Theorem dfixp 8927

Description: Eliminate the expression {𝑥 ∣ 𝑥 ∈ 𝐴} in df-ixp 8926, 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 8926	. 2 ⊢ X𝑥 ∈ 𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)}
2		abid2 2898	. . . . 5 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
3	2	fneq2i 6637	. . . 4 ⊢ (𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ 𝑓 Fn 𝐴)
4	3	anbi1i 636	. . 3 ⊢ ((𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵) ↔ (𝑓 Fn 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵))
5	4	abbii 2828	. 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 2784	1 ⊢ X𝑥 ∈ 𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)}

This proof depends on syntax axioms:   ∧ wa 401   = wceq 1570   ∈ wcel 2145  {cab 2739  ∀wral 3077   Fn wfn 6533  ‘cfv 6538  Xcixp 8925

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-fn 6541  df-ixp 8926

This theorem is used by:  ixpsnval  8928  elixp2  8929  ixpeq1  8936  cbvixp  8942  cbvixpv  8943  ixp0x  8954