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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 3908 | . . . . . 6 ⊢ (𝑥 ⊆ 𝐴 → (𝐴 ⊆ 𝐵 → 𝑥 ⊆ 𝐵)) | |
2 | 1 | com12 32 | . . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ⊆ 𝐴 → 𝑥 ⊆ 𝐵)) |
3 | 2 | anim1d 614 | . . . 4 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → (𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅))) |
4 | 3 | imim1d 82 | . . 3 ⊢ (𝐴 ⊆ 𝐵 → (((𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦) → ((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))) |
5 | 4 | alimdv 1924 | . 2 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑥((𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦) → ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))) |
6 | df-fr 5509 | . 2 ⊢ (𝑅 Fr 𝐵 ↔ ∀𝑥((𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦)) | |
7 | df-fr 5509 | . 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦)) | |
8 | 5, 6, 7 | 3imtr4g 299 | 1 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Fr 𝐵 → 𝑅 Fr 𝐴)) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 399 ∀wal 1541 ≠ wne 2940 ∀wral 3061 ∃wrex 3062 ⊆ wss 3866 ∅c0 4237 class class class wbr 5053 Fr wfr 5506 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1803 ax-4 1817 ax-5 1918 ax-6 1976 ax-7 2016 ax-8 2112 ax-9 2120 ax-ext 2708 |
This theorem depends on definitions: df-bi 210 df-an 400 df-tru 1546 df-ex 1788 df-sb 2071 df-clab 2715 df-cleq 2729 df-clel 2816 df-v 3410 df-in 3873 df-ss 3883 df-fr 5509 |
This theorem is referenced by: freq2 5522 wess 5538 fprlem1 8041 frmin 9365 frrlem15 9373 |
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