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Mirrors > Home > MPE Home > Th. List > frss | Structured version Visualization version GIF version |
Description: Subset theorem for the well-founded predicate. Exercise 1 of [TakeutiZaring] p. 31. (Contributed by NM, 3-Apr-1994.) (Proof shortened by Andrew Salmon, 25-Jul-2011.) |
Ref | Expression |
---|---|
frss | ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Fr 𝐵 → 𝑅 Fr 𝐴)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | sstr2 3575 | . . . . . 6 ⊢ (𝑥 ⊆ 𝐴 → (𝐴 ⊆ 𝐵 → 𝑥 ⊆ 𝐵)) | |
2 | 1 | com12 32 | . . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ⊆ 𝐴 → 𝑥 ⊆ 𝐵)) |
3 | 2 | anim1d 586 | . . . 4 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → (𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅))) |
4 | 3 | imim1d 80 | . . 3 ⊢ (𝐴 ⊆ 𝐵 → (((𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦) → ((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))) |
5 | 4 | alimdv 1832 | . 2 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑥((𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦) → ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))) |
6 | df-fr 4997 | . 2 ⊢ (𝑅 Fr 𝐵 ↔ ∀𝑥((𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦)) | |
7 | df-fr 4997 | . 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦)) | |
8 | 5, 6, 7 | 3imtr4g 284 | 1 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Fr 𝐵 → 𝑅 Fr 𝐴)) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 383 ∀wal 1473 ≠ wne 2780 ∀wral 2896 ∃wrex 2897 ⊆ wss 3540 ∅c0 3874 class class class wbr 4583 Fr wfr 4994 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 |
This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-clab 2597 df-cleq 2603 df-clel 2606 df-in 3547 df-ss 3554 df-fr 4997 |
This theorem is referenced by: freq2 5009 wess 5025 frmin 30983 frrlem5 31028 |
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