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Mirrors > Home > ILE Home > Th. List > Mathboxes > setindft | GIF version |
Description: Axiom of set-induction with a disjoint variable condition replaced with a nonfreeness hypothesis. (Contributed by BJ, 22-Nov-2019.) |
Ref | Expression |
---|---|
setindft | ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑥(∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → ∀𝑥𝜑)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | nfa1 1521 | . . 3 ⊢ Ⅎ𝑥∀𝑥Ⅎ𝑦𝜑 | |
2 | nfv 1508 | . . . . . 6 ⊢ Ⅎ𝑧∀𝑥Ⅎ𝑦𝜑 | |
3 | nfnf1 1524 | . . . . . . 7 ⊢ Ⅎ𝑦Ⅎ𝑦𝜑 | |
4 | 3 | nfal 1556 | . . . . . 6 ⊢ Ⅎ𝑦∀𝑥Ⅎ𝑦𝜑 |
5 | nfsbt 1956 | . . . . . 6 ⊢ (∀𝑥Ⅎ𝑦𝜑 → Ⅎ𝑦[𝑧 / 𝑥]𝜑) | |
6 | nfv 1508 | . . . . . . 7 ⊢ Ⅎ𝑧[𝑦 / 𝑥]𝜑 | |
7 | 6 | a1i 9 | . . . . . 6 ⊢ (∀𝑥Ⅎ𝑦𝜑 → Ⅎ𝑧[𝑦 / 𝑥]𝜑) |
8 | sbequ 1820 | . . . . . . 7 ⊢ (𝑧 = 𝑦 → ([𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)) | |
9 | 8 | a1i 9 | . . . . . 6 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (𝑧 = 𝑦 → ([𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑))) |
10 | 2, 4, 5, 7, 9 | cbvrald 13373 | . . . . 5 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 ↔ ∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑)) |
11 | 10 | biimpd 143 | . . . 4 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 → ∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑)) |
12 | 11 | imim1d 75 | . . 3 ⊢ (∀𝑥Ⅎ𝑦𝜑 → ((∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → (∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 → 𝜑))) |
13 | 1, 12 | alimd 1501 | . 2 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑥(∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → ∀𝑥(∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 → 𝜑))) |
14 | ax-setind 4496 | . 2 ⊢ (∀𝑥(∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 → 𝜑) → ∀𝑥𝜑) | |
15 | 13, 14 | syl6 33 | 1 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑥(∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → ∀𝑥𝜑)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ↔ wb 104 ∀wal 1333 Ⅎwnf 1440 [wsb 1742 ∀wral 2435 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-io 699 ax-5 1427 ax-7 1428 ax-gen 1429 ax-ie1 1473 ax-ie2 1474 ax-8 1484 ax-10 1485 ax-11 1486 ax-i12 1487 ax-bndl 1489 ax-4 1490 ax-17 1506 ax-i9 1510 ax-ial 1514 ax-i5r 1515 ax-ext 2139 ax-setind 4496 |
This theorem depends on definitions: df-bi 116 df-nf 1441 df-sb 1743 df-cleq 2150 df-clel 2153 df-ral 2440 |
This theorem is referenced by: setindf 13552 |
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