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| Mirrors > Home > ILE Home > Th. List > Mathboxes > alseud | GIF version | ||
| Description: Introduction rule: "all some one" holds if the "for all" part holds and the antecedent has exactly one witness. This is the converse of alseu1d 17291 and alseu2d 17292 taken together. (Contributed by David A. Wheeler, 22-Jul-2026.) |
| Ref | Expression |
|---|---|
| alseud.1 | ⊢ (𝜑 → ∀𝑥(𝜓 → 𝜒)) |
| alseud.2 | ⊢ (𝜑 → ∃!𝑥𝜓) |
| Ref | Expression |
|---|---|
| alseud | ⊢ (𝜑 → ∀∃!𝑥(𝜓 → 𝜒)) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | alseud.1 | . 2 ⊢ (𝜑 → ∀𝑥(𝜓 → 𝜒)) | |
| 2 | alseud.2 | . 2 ⊢ (𝜑 → ∃!𝑥𝜓) | |
| 3 | df-alseu 17284 | . 2 ⊢ (∀∃!𝑥(𝜓 → 𝜒) ↔ (∀𝑥(𝜓 → 𝜒) ∧ ∃!𝑥𝜓)) | |
| 4 | 1, 2, 3 | sylanbrc 421 | 1 ⊢ (𝜑 → ∀∃!𝑥(𝜓 → 𝜒)) |
| Colors of variables: wff set class |
| This proof depends on syntax axioms: → wi 4 ∀wal 1400 ∃!weu 2086 ∀∃!walseu 17282 |
| This proof depends on axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 |
| This proof depends on definitions: df-bi 117 df-alseu 17284 |
| This theorem is used by: (None) |
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