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Mirrors > Home > MPE Home > Th. List > Mathboxes > alsi-no-surprise | Structured version Visualization version GIF version |
Description: Demonstrate that there is never a "surprise" when using the allsome quantifier, that is, it is never possible for the consequent to be both always true and always false. This uses the definition of df-alsi 46492; the proof itself builds on alimp-no-surprise 46485. For a contrast, see alimp-surprise 46484. (Contributed by David A. Wheeler, 27-Oct-2018.) |
Ref | Expression |
---|---|
alsi-no-surprise | ⊢ ¬ (∀!𝑥(𝜑 → 𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | alimp-no-surprise 46485 | . 2 ⊢ ¬ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑) | |
2 | df-alsi 46492 | . . . 4 ⊢ (∀!𝑥(𝜑 → 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑)) | |
3 | df-alsi 46492 | . . . 4 ⊢ (∀!𝑥(𝜑 → ¬ 𝜓) ↔ (∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)) | |
4 | 2, 3 | anbi12i 627 | . . 3 ⊢ ((∀!𝑥(𝜑 → 𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓)) ↔ ((∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑) ∧ (∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))) |
5 | anandi3r 1102 | . . 3 ⊢ ((∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑 ∧ ∀𝑥(𝜑 → ¬ 𝜓)) ↔ ((∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑) ∧ (∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))) | |
6 | 3ancomb 1098 | . . 3 ⊢ ((∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑 ∧ ∀𝑥(𝜑 → ¬ 𝜓)) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)) | |
7 | 4, 5, 6 | 3bitr2i 299 | . 2 ⊢ ((∀!𝑥(𝜑 → 𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓)) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)) |
8 | 1, 7 | mtbir 323 | 1 ⊢ ¬ (∀!𝑥(𝜑 → 𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓)) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 396 ∧ w3a 1086 ∀wal 1537 ∃wex 1782 ∀!walsi 46490 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1798 ax-4 1812 |
This theorem depends on definitions: df-bi 206 df-an 397 df-3an 1088 df-ex 1783 df-alsi 46492 |
This theorem is referenced by: (None) |
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