Home | Metamath
Proof Explorer Theorem List (p. 459 of 459) | < 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
(1-28814) |
Hilbert Space Explorer
(28815-30337) |
Users' Mathboxes
(30338-45830) |
Type | Label | Description |
---|---|---|
Statement | ||
Theorem | mvlrmuld 45801 | Move the right term in a product on the LHS to the RHS, deduction form. (Contributed by David A. Wheeler, 11-Oct-2018.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ≠ 0) & ⊢ (𝜑 → (𝐴 · 𝐵) = 𝐶) ⇒ ⊢ (𝜑 → 𝐴 = (𝐶 / 𝐵)) | ||
Theorem | mvlrmuli 45802 | Move the right term in a product on the LHS to the RHS, inference form. (Contributed by David A. Wheeler, 11-Oct-2018.) |
⊢ 𝐴 ∈ ℂ & ⊢ 𝐵 ∈ ℂ & ⊢ 𝐵 ≠ 0 & ⊢ (𝐴 · 𝐵) = 𝐶 ⇒ ⊢ 𝐴 = (𝐶 / 𝐵) | ||
Examples using the algebra helpers. | ||
Theorem | i2linesi 45803 | Solve for the intersection of two lines expressed in Y = MX+B form (note that the lines cannot be vertical). Here we use inference form. We just solve for X, since Y can be trivially found by using X. This is an example of how to use the algebra helpers. Notice that because this proof uses algebra helpers, the main steps of the proof are higher level and easier to follow by a human reader. (Contributed by David A. Wheeler, 11-Oct-2018.) |
⊢ 𝐴 ∈ ℂ & ⊢ 𝐵 ∈ ℂ & ⊢ 𝐶 ∈ ℂ & ⊢ 𝐷 ∈ ℂ & ⊢ 𝑋 ∈ ℂ & ⊢ 𝑌 = ((𝐴 · 𝑋) + 𝐵) & ⊢ 𝑌 = ((𝐶 · 𝑋) + 𝐷) & ⊢ (𝐴 − 𝐶) ≠ 0 ⇒ ⊢ 𝑋 = ((𝐷 − 𝐵) / (𝐴 − 𝐶)) | ||
Theorem | i2linesd 45804 | Solve for the intersection of two lines expressed in Y = MX+B form (note that the lines cannot be vertical). Here we use deduction form. We just solve for X, since Y can be trivially found by using X. This is an example of how to use the algebra helpers. Notice that because this proof uses algebra helpers, the main steps of the proof are higher level and easier to follow by a human reader. (Contributed by David A. Wheeler, 15-Oct-2018.) |
⊢ (𝜑 → 𝐴 ∈ ℂ) & ⊢ (𝜑 → 𝐵 ∈ ℂ) & ⊢ (𝜑 → 𝐶 ∈ ℂ) & ⊢ (𝜑 → 𝐷 ∈ ℂ) & ⊢ (𝜑 → 𝑋 ∈ ℂ) & ⊢ (𝜑 → 𝑌 = ((𝐴 · 𝑋) + 𝐵)) & ⊢ (𝜑 → 𝑌 = ((𝐶 · 𝑋) + 𝐷)) & ⊢ (𝜑 → (𝐴 − 𝐶) ≠ 0) ⇒ ⊢ (𝜑 → 𝑋 = ((𝐷 − 𝐵) / (𝐴 − 𝐶))) | ||
Prove that some formal expressions using classical logic have meanings that might not be obvious to some lay readers. I find these are common mistakes and are worth pointing out to new people. In particular we prove alimp-surprise 45805, empty-surprise 45807, and eximp-surprise 45809. | ||
Theorem | alimp-surprise 45805 |
Demonstrate that when using "for all" and material implication the
consequent can be both always true and always false if there is no case
where the antecedent is true.
Those inexperienced with formal notations of classical logic can be surprised with what "for all" and material implication do together when the implication's antecedent is never true. This can happen, for example, when the antecedent is set membership but the set is the empty set (e.g., 𝑥 ∈ 𝑀 and 𝑀 = ∅). This is perhaps best explained using an example. The sentence "All Martians are green" would typically be represented formally using the expression ∀𝑥(𝜑 → 𝜓). In this expression 𝜑 is true iff 𝑥 is a Martian and 𝜓 is true iff 𝑥 is green. Similarly, "All Martians are not green" would typically be represented as ∀𝑥(𝜑 → ¬ 𝜓). However, if there are no Martians (¬ ∃𝑥𝜑), then both of those expressions are true. That is surprising to the inexperienced, because the two expressions seem to be the opposite of each other. The reason this occurs is because in classical logic the implication (𝜑 → 𝜓) is equivalent to ¬ 𝜑 ∨ 𝜓 (as proven in imor 850). When 𝜑 is always false, ¬ 𝜑 is always true, and an or with true is always true. Here are a few technical notes. In this notation, 𝜑 and 𝜓 are predicates that return a true or false value and may depend on 𝑥. We only say may because it actually doesn't matter for our proof. In Metamath this simply means that we do not require that 𝜑, 𝜓, and 𝑥 be distinct (so 𝑥 can be part of 𝜑 or 𝜓). In natural language the term "implies" often presumes that the antecedent can occur in at one least circumstance and that there is some sort of causality. However, exactly what causality means is complex and situation-dependent. Modern logic typically uses material implication instead; this has a rigorous definition, but it is important for new users of formal notation to precisely understand it. There are ways to solve this, e.g., expressly stating that the antecedent exists (see alimp-no-surprise 45806) or using the allsome quantifier (df-alsi 45813) . For other "surprises" for new users of classical logic, see empty-surprise 45807 and eximp-surprise 45809. (Contributed by David A. Wheeler, 17-Oct-2018.) |
⊢ ¬ ∃𝑥𝜑 ⇒ ⊢ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓)) | ||
Theorem | alimp-no-surprise 45806 | There is no "surprise" in a for-all with implication if there exists a value where the antecedent is true. This is one way to prevent for-all with implication from allowing anything. For a contrast, see alimp-surprise 45805. The allsome quantifier also counters this problem, see df-alsi 45813. (Contributed by David A. Wheeler, 27-Oct-2018.) |
⊢ ¬ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑) | ||
Theorem | empty-surprise 45807 |
Demonstrate that when using restricted "for all" over a class the
expression can be both always true and always false if the class is
empty.
Those inexperienced with formal notations of classical logic can be surprised with what restricted "for all" does over an empty set. It is important to note that ∀𝑥 ∈ 𝐴𝜑 is simply an abbreviation for ∀𝑥(𝑥 ∈ 𝐴 → 𝜑) (per df-ral 3075). Thus, if 𝐴 is the empty set, this expression is always true regardless of the value of 𝜑 (see alimp-surprise 45805). If you want the expression ∀𝑥 ∈ 𝐴𝜑 to not be vacuously true, you need to ensure that set 𝐴 is inhabited (e.g., ∃𝑥 ∈ 𝐴). (Technical note: You can also assert that 𝐴 ≠ ∅; this is an equivalent claim in classical logic as proven in n0 4247, but in intuitionistic logic the statement 𝐴 ≠ ∅ is a weaker claim than ∃𝑥 ∈ 𝐴.) Some materials on logic (particularly those that discuss "syllogisms") are based on the much older work by Aristotle, but Aristotle expressly excluded empty sets from his system. Aristotle had a specific goal; he was trying to develop a "companion-logic" for science. He relegates fictions like fairy godmothers and mermaids and unicorns to the realms of poetry and literature... This is why he leaves no room for such nonexistent entities in his logic." (Groarke, "Aristotle: Logic", section 7. (Existential Assumptions), Internet Encyclopedia of Philosophy, http://www.iep.utm.edu/aris-log/ 4247). While this made sense for his purposes, it is less flexible than modern (classical) logic which does permit empty sets. If you wish to make claims that require a nonempty set, you must expressly include that requirement, e.g., by stating ∃𝑥𝜑. Examples of proofs that do this include barbari 2690, celaront 2692, and cesaro 2699. For another "surprise" for new users of classical logic, see alimp-surprise 45805 and eximp-surprise 45809. (Contributed by David A. Wheeler, 20-Oct-2018.) |
⊢ ¬ ∃𝑥 𝑥 ∈ 𝐴 ⇒ ⊢ ∀𝑥 ∈ 𝐴 𝜑 | ||
Theorem | empty-surprise2 45808 |
"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 45807. Here we prove an extreme example: we "prove" that false is true. Of course, we actually do no such thing (see notfal 1566); 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 45814. (Contributed by David A. Wheeler, 20-Oct-2018.) |
⊢ ¬ ∃𝑥 𝑥 ∈ 𝐴 ⇒ ⊢ ∀𝑥 ∈ 𝐴 ⊥ | ||
Theorem | eximp-surprise 45809 |
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 850, 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 45810. See also alimp-surprise 45805 and empty-surprise 45807. (Contributed by David A. Wheeler, 17-Oct-2018.) |
⊢ (∃𝑥(𝜑 → 𝜓) ↔ ∃𝑥(¬ 𝜑 ∨ 𝜓)) | ||
Theorem | eximp-surprise2 45810 |
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 45809, which shows what implication really expands to. See also empty-surprise 45807. (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 45805 and empty-surprise 45807 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 45807 I hope that others will eventually agree that allsome is awesome. | ||
Syntax | walsi 45811 | 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 45812 | 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 45813 | 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 45814 | 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 45815 | 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 45816 | 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 45817 | 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 45818 | 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 45819 | 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 45820* | 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 45821 | 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 45813; the proof itself builds on alimp-no-surprise 45806. For a contrast, see alimp-surprise 45805. (Contributed by David A. Wheeler, 27-Oct-2018.) |
⊢ ¬ (∀!𝑥(𝜑 → 𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓)) | ||
Miscellaneous proofs. | ||
Theorem | 5m4e1 45822 | Prove that 5 - 4 = 1. (Contributed by David A. Wheeler, 31-Jan-2017.) |
⊢ (5 − 4) = 1 | ||
Theorem | 2p2ne5 45823 | 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 45824 | 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 45825 | 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 45826* | 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 45827 | Weighted version of amgmlem 25688. (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 45828 | Alternate proof of amgmlem 25688 using amgmwlem 45827. (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 45829 | Weighted arithmetic-geometric mean inequality for 𝑛 = 2 (compare amgm2d 41322). (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 45830 | Young's inequality for 𝑛 = 2, a direct application of amgmw2d 45829. (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 > |