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| Mirrors > Home > ILE Home > Th. List > dfixp | GIF version | ||
| Description: Eliminate the expression {𝑥 ∣ 𝑥 ∈ 𝐴} in df-ixp 6844, 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 6844 | . 2 ⊢ X𝑥 ∈ 𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn {𝑥 ∣ 𝑥 ∈ 𝐴} ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)} | |
| 2 | abid2 2350 | . . . . 5 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴 | |
| 3 | 2 | fneq2i 5415 | . . . 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 2345 | . 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 2250 | 1 ⊢ X𝑥 ∈ 𝐴 𝐵 = {𝑓 ∣ (𝑓 Fn 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝑓‘𝑥) ∈ 𝐵)} |
| Colors of variables: wff set class |
| Syntax hints: ∧ wa 104 = wceq 1395 ∈ wcel 2200 {cab 2215 ∀wral 2508 Fn wfn 5312 ‘cfv 5317 Xcixp 6843 |
| 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 1493 ax-7 1494 ax-gen 1495 ax-ie1 1539 ax-ie2 1540 ax-8 1550 ax-11 1552 ax-4 1556 ax-17 1572 ax-i9 1576 ax-ial 1580 ax-i5r 1581 ax-ext 2211 |
| This theorem depends on definitions: df-bi 117 df-tru 1398 df-nf 1507 df-sb 1809 df-clab 2216 df-cleq 2222 df-clel 2225 df-fn 5320 df-ixp 6844 |
| This theorem is referenced by: ixpsnval 6846 elixp2 6847 ixpeq1 6854 cbvixp 6860 ixp0x 6871 |
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