Home | Metamath
Proof Explorer Theorem List (p. 462 of 462) | < Previous Wrap > |
Bad symbols? Try the
GIF version. |
||
Mirrors > Metamath Home Page > MPE Home Page > Theorem List Contents > Recent Proofs This page: Page List |
Color key: | Metamath Proof Explorer
(1-28953) |
Hilbert Space Explorer
(28954-30476) |
Users' Mathboxes
(30477-46123) |
Type | Label | Description |
---|---|---|
Statement | ||
Theorem | empty-surprise2 46101 |
"Prove" that false is true when using a restricted "for
all" over the
empty set, to demonstrate that the expression is always true if the
value ranges over the empty set.
Those inexperienced with formal notations of classical logic can be surprised with what restricted "for all" does over an empty set. We proved the general case in empty-surprise 46100. Here we prove an extreme example: we "prove" that false is true. Of course, we actually do no such thing (see notfal 1571); the problem is that restricted "for all" works in ways that might seem counterintuitive to the inexperienced when given an empty set. Solutions to this can include requiring that the set not be empty or by using the allsome quantifier df-alsc 46107. (Contributed by David A. Wheeler, 20-Oct-2018.) |
⊢ ¬ ∃𝑥 𝑥 ∈ 𝐴 ⇒ ⊢ ∀𝑥 ∈ 𝐴 ⊥ | ||
Theorem | eximp-surprise 46102 |
Show what implication inside "there exists" really expands to (using
implication directly inside "there exists" is usually a
mistake).
Those inexperienced with formal notations of classical logic may use expressions combining "there exists" with implication. That is usually a mistake, because as proven using imor 853, such an expression can be rewritten using not with or - and that is often not what the author intended. New users of formal notation who use "there exists" with an implication should consider if they meant "and" instead of "implies". A stark example is shown in eximp-surprise2 46103. See also alimp-surprise 46098 and empty-surprise 46100. (Contributed by David A. Wheeler, 17-Oct-2018.) |
⊢ (∃𝑥(𝜑 → 𝜓) ↔ ∃𝑥(¬ 𝜑 ∨ 𝜓)) | ||
Theorem | eximp-surprise2 46103 |
Show that "there exists" with an implication is always true if there
exists a situation where the antecedent is false.
Those inexperienced with formal notations of classical logic may use expressions combining "there exists" with implication. This is usually a mistake, because that combination does not mean what an inexperienced person might think it means. For example, if there is some object that does not meet the precondition 𝜑, then the expression ∃𝑥(𝜑 → 𝜓) as a whole is always true, no matter what 𝜓 is (𝜓 could even be false, ⊥). New users of formal notation who use "there exists" with an implication should consider if they meant "and" instead of "implies". See eximp-surprise 46102, which shows what implication really expands to. See also empty-surprise 46100. (Contributed by David A. Wheeler, 18-Oct-2018.) |
⊢ ∃𝑥 ¬ 𝜑 ⇒ ⊢ ∃𝑥(𝜑 → 𝜓) | ||
These are definitions and proofs involving an experimental "allsome" quantifier (aka "all some"). In informal language, statements like "All Martians are green" imply that there is at least one Martian. But it's easy to mistranslate informal language into formal notations because similar statements like ∀𝑥𝜑 → 𝜓 do not imply that 𝜑 is ever true, leading to vacuous truths. See alimp-surprise 46098 and empty-surprise 46100 as examples of the problem. Some systems include a mechanism to counter this, e.g., PVS allows types to be appended with "+" to declare that they are nonempty. This section presents a different solution to the same problem. The "allsome" quantifier expressly includes the notion of both "all" and "there exists at least one" (aka some), and is defined to make it easier to more directly express both notions. The hope is that if a quantifier more directly expresses this concept, it will be used instead and reduce the risk of creating formal expressions that look okay but in fact are mistranslations. The term "allsome" was chosen because it's short, easy to say, and clearly hints at the two concepts it combines. I do not expect this to be used much in Metamath, because in Metamath there's a general policy of avoiding the use of new definitions unless there are very strong reasons to do so. Instead, my goal is to rigorously define this quantifier and demonstrate a few basic properties of it. The syntax allows two forms that look like they would be problematic, but they are fine. When applied to a top-level implication we allow ∀!𝑥(𝜑 → 𝜓), and when restricted (applied to a class) we allow ∀!𝑥 ∈ 𝐴𝜑. The first symbol after the setvar variable must always be ∈ if it is the form applied to a class, and since ∈ cannot begin a wff, it is unambiguous. The → looks like it would be a problem because 𝜑 or 𝜓 might include implications, but any implication arrow → within any wff must be surrounded by parentheses, so only the implication arrow of ∀! can follow the wff. The implication syntax would work fine without the parentheses, but I added the parentheses because it makes things clearer inside larger complex expressions, and it's also more consistent with the rest of the syntax. For more, see "The Allsome Quantifier" by David A. Wheeler at https://dwheeler.com/essays/allsome.html 46100 I hope that others will eventually agree that allsome is awesome. | ||
Syntax | walsi 46104 | Extend wff definition to include "all some" applied to a top-level implication, which means 𝜓 is true whenever 𝜑 is true, and there is at least least one 𝑥 where 𝜑 is true. (Contributed by David A. Wheeler, 20-Oct-2018.) |
wff ∀!𝑥(𝜑 → 𝜓) | ||
Syntax | walsc 46105 | Extend wff definition to include "all some" applied to a class, which means 𝜑 is true for all 𝑥 in 𝐴, and there is at least one 𝑥 in 𝐴. (Contributed by David A. Wheeler, 20-Oct-2018.) |
wff ∀!𝑥 ∈ 𝐴𝜑 | ||
Definition | df-alsi 46106 | Define "all some" applied to a top-level implication, which means 𝜓 is true whenever 𝜑 is true and there is at least one 𝑥 where 𝜑 is true. (Contributed by David A. Wheeler, 20-Oct-2018.) |
⊢ (∀!𝑥(𝜑 → 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑)) | ||
Definition | df-alsc 46107 | Define "all some" applied to a class, which means 𝜑 is true for all 𝑥 in 𝐴 and there is at least one 𝑥 in 𝐴. (Contributed by David A. Wheeler, 20-Oct-2018.) |
⊢ (∀!𝑥 ∈ 𝐴𝜑 ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑥 𝑥 ∈ 𝐴)) | ||
Theorem | alsconv 46108 | There is an equivalence between the two "all some" forms. (Contributed by David A. Wheeler, 22-Oct-2018.) |
⊢ (∀!𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀!𝑥 ∈ 𝐴𝜑) | ||
Theorem | alsi1d 46109 | Deduction rule: Given "all some" applied to a top-level inference, you can extract the "for all" part. (Contributed by David A. Wheeler, 20-Oct-2018.) |
⊢ (𝜑 → ∀!𝑥(𝜓 → 𝜒)) ⇒ ⊢ (𝜑 → ∀𝑥(𝜓 → 𝜒)) | ||
Theorem | alsi2d 46110 | Deduction rule: Given "all some" applied to a top-level inference, you can extract the "exists" part. (Contributed by David A. Wheeler, 20-Oct-2018.) |
⊢ (𝜑 → ∀!𝑥(𝜓 → 𝜒)) ⇒ ⊢ (𝜑 → ∃𝑥𝜓) | ||
Theorem | alsc1d 46111 | Deduction rule: Given "all some" applied to a class, you can extract the "for all" part. (Contributed by David A. Wheeler, 20-Oct-2018.) |
⊢ (𝜑 → ∀!𝑥 ∈ 𝐴𝜓) ⇒ ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 𝜓) | ||
Theorem | alsc2d 46112 | Deduction rule: Given "all some" applied to a class, you can extract the "there exists" part. (Contributed by David A. Wheeler, 20-Oct-2018.) |
⊢ (𝜑 → ∀!𝑥 ∈ 𝐴𝜓) ⇒ ⊢ (𝜑 → ∃𝑥 𝑥 ∈ 𝐴) | ||
Theorem | alscn0d 46113* | Deduction rule: Given "all some" applied to a class, the class is not the empty set. (Contributed by David A. Wheeler, 23-Oct-2018.) |
⊢ (𝜑 → ∀!𝑥 ∈ 𝐴𝜓) ⇒ ⊢ (𝜑 → 𝐴 ≠ ∅) | ||
Theorem | alsi-no-surprise 46114 | 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 46106; the proof itself builds on alimp-no-surprise 46099. For a contrast, see alimp-surprise 46098. (Contributed by David A. Wheeler, 27-Oct-2018.) |
⊢ ¬ (∀!𝑥(𝜑 → 𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓)) | ||
Miscellaneous proofs. | ||
Theorem | 5m4e1 46115 | Prove that 5 - 4 = 1. (Contributed by David A. Wheeler, 31-Jan-2017.) |
⊢ (5 − 4) = 1 | ||
Theorem | 2p2ne5 46116 | Prove that 2 + 2 ≠ 5. In George Orwell's "1984", Part One, Chapter Seven, the protagonist Winston notes that, "In the end the Party would announce that two and two made five, and you would have to believe it." http://www.sparknotes.com/lit/1984/section4.rhtml. More generally, the phrase 2 + 2 = 5 has come to represent an obviously false dogma one may be required to believe. See the Wikipedia article for more about this: https://en.wikipedia.org/wiki/2_%2B_2_%3D_5. Unsurprisingly, we can easily prove that this claim is false. (Contributed by David A. Wheeler, 31-Jan-2017.) |
⊢ (2 + 2) ≠ 5 | ||
Theorem | resolution 46117 | Resolution rule. This is the primary inference rule in some automated theorem provers such as prover9. The resolution rule can be traced back to Davis and Putnam (1960). (Contributed by David A. Wheeler, 9-Feb-2017.) |
⊢ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜒)) → (𝜓 ∨ 𝜒)) | ||
Theorem | testable 46118 | In classical logic all wffs are testable, that is, it is always true that (¬ 𝜑 ∨ ¬ ¬ 𝜑). This is not necessarily true in intuitionistic logic. In intuitionistic logic, if this statement is true for some 𝜑, then 𝜑 is testable. The proof is trivial because it's simply a special case of the law of the excluded middle, which is true in classical logic but not necessarily true in intuitionisic logic. (Contributed by David A. Wheeler, 5-Dec-2018.) |
⊢ (¬ 𝜑 ∨ ¬ ¬ 𝜑) | ||
Theorem | aacllem 46119* | Lemma for other theorems about 𝔸. (Contributed by Brendan Leahy, 3-Jan-2020.) (Revised by Alexander van der Vekens and David A. Wheeler, 25-Apr-2020.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝑁 ∈ ℕ0) & ⊢ ((𝜑 ∧ 𝑛 ∈ (1...𝑁)) → 𝑋 ∈ ℂ) & ⊢ ((𝜑 ∧ 𝑘 ∈ (0...𝑁) ∧ 𝑛 ∈ (1...𝑁)) → 𝐶 ∈ ℚ) & ⊢ ((𝜑 ∧ 𝑘 ∈ (0...𝑁)) → (𝐴↑𝑘) = Σ𝑛 ∈ (1...𝑁)(𝐶 · 𝑋)) ⇒ ⊢ (𝜑 → 𝐴 ∈ 𝔸) | ||
Theorem | amgmwlem 46120 | Weighted version of amgmlem 25826. (Contributed by Kunhao Zheng, 19-Jun-2021.) |
⊢ 𝑀 = (mulGrp‘ℂfld) & ⊢ (𝜑 → 𝐴 ∈ Fin) & ⊢ (𝜑 → 𝐴 ≠ ∅) & ⊢ (𝜑 → 𝐹:𝐴⟶ℝ+) & ⊢ (𝜑 → 𝑊:𝐴⟶ℝ+) & ⊢ (𝜑 → (ℂfld Σg 𝑊) = 1) ⇒ ⊢ (𝜑 → (𝑀 Σg (𝐹 ∘f ↑𝑐𝑊)) ≤ (ℂfld Σg (𝐹 ∘f · 𝑊))) | ||
Theorem | amgmlemALT 46121 | Alternate proof of amgmlem 25826 using amgmwlem 46120. (Contributed by Kunhao Zheng, 20-Jun-2021.) (Proof modification is discouraged.) (New usage is discouraged.) |
⊢ 𝑀 = (mulGrp‘ℂfld) & ⊢ (𝜑 → 𝐴 ∈ Fin) & ⊢ (𝜑 → 𝐴 ≠ ∅) & ⊢ (𝜑 → 𝐹:𝐴⟶ℝ+) ⇒ ⊢ (𝜑 → ((𝑀 Σg 𝐹)↑𝑐(1 / (♯‘𝐴))) ≤ ((ℂfld Σg 𝐹) / (♯‘𝐴))) | ||
Theorem | amgmw2d 46122 | Weighted arithmetic-geometric mean inequality for 𝑛 = 2 (compare amgm2d 41428). (Contributed by Kunhao Zheng, 20-Jun-2021.) |
⊢ (𝜑 → 𝐴 ∈ ℝ+) & ⊢ (𝜑 → 𝑃 ∈ ℝ+) & ⊢ (𝜑 → 𝐵 ∈ ℝ+) & ⊢ (𝜑 → 𝑄 ∈ ℝ+) & ⊢ (𝜑 → (𝑃 + 𝑄) = 1) ⇒ ⊢ (𝜑 → ((𝐴↑𝑐𝑃) · (𝐵↑𝑐𝑄)) ≤ ((𝐴 · 𝑃) + (𝐵 · 𝑄))) | ||
Theorem | young2d 46123 | Young's inequality for 𝑛 = 2, a direct application of amgmw2d 46122. (Contributed by Kunhao Zheng, 6-Jul-2021.) |
⊢ (𝜑 → 𝐴 ∈ ℝ+) & ⊢ (𝜑 → 𝑃 ∈ ℝ+) & ⊢ (𝜑 → 𝐵 ∈ ℝ+) & ⊢ (𝜑 → 𝑄 ∈ ℝ+) & ⊢ (𝜑 → ((1 / 𝑃) + (1 / 𝑄)) = 1) ⇒ ⊢ (𝜑 → (𝐴 · 𝐵) ≤ (((𝐴↑𝑐𝑃) / 𝑃) + ((𝐵↑𝑐𝑄) / 𝑄))) |
< Previous Wrap > |
Copyright terms: Public domain | < Previous Wrap > |