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| Description: Eliminate the expression {𝑥 ∣ 𝑥 ∈ 𝐴} in df-ixp 8938, 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.) | 
| Ref | Expression | 
|---|---|
| dfixp | ⊢ X𝑥 ∈ 𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)} | 
| Step | Hyp | Ref | Expression | 
|---|---|---|---|
| 1 | df-ixp 8938 | . 2 ⊢ X𝑥 ∈ 𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)} | |
| 2 | abid2 2879 | . . . . 5 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴 | |
| 3 | 2 | fneq2i 6666 | . . . 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 2809 | . 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 2765 | 1 ⊢ X𝑥 ∈ 𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)} | 
| Colors of variables: wff setvar class | 
| Syntax hints: ∧ wa 395 = wceq 1540 ∈ wcel 2108 {cab 2714 ∀wral 3061 Fn wfn 6556 ‘cfv 6561 Xcixp 8937 | 
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1795 ax-4 1809 ax-5 1910 ax-6 1967 ax-7 2007 ax-8 2110 ax-9 2118 ax-ext 2708 | 
| This theorem depends on definitions: df-bi 207 df-an 396 df-tru 1543 df-ex 1780 df-sb 2065 df-clab 2715 df-cleq 2729 df-clel 2816 df-fn 6564 df-ixp 8938 | 
| This theorem is referenced by: ixpsnval 8940 elixp2 8941 ixpeq1 8948 cbvixp 8954 cbvixpv 8955 ixp0x 8966 | 
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