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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 17143 and alseu2d 17144 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 17136 | . 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 |
| Syntax hints: → wi 4 ∀wal 1400 ∃!weu 2086 ∀∃!walseu 17134 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 |
| This theorem depends on definitions: df-bi 117 df-alseu 17136 |
| This theorem is referenced by: (None) |
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