P. 503 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 50201-50238   *Has distinct variable group(s)

Type	Label	Description
Statement
Definition	df-reflexive 50201*	Define reflexive relation; relation 𝑅 is reflexive over the set 𝐴 iff ∀𝑥 ∈ 𝐴𝑥𝑅𝑥. (Contributed by David A. Wheeler, 1-Dec-2019.)
⊢ (𝑅Reflexive𝐴 ↔ (𝑅 ⊆ (𝐴 × 𝐴) ∧ ∀𝑥 ∈ 𝐴 𝑥𝑅𝑥))
Syntax	wirreflexive 50202	Extend wff definition to include "Irreflexive" applied to a class, which is true iff class R is an irreflexive relation over the set A. See df-irreflexive 50203. (Contributed by David A. Wheeler, 1-Dec-2019.)
wff 𝑅Irreflexive𝐴
Definition	df-irreflexive 50203*	Define irreflexive relation; relation 𝑅 is irreflexive over the set 𝐴 iff ∀𝑥 ∈ 𝐴¬ 𝑥𝑅𝑥. Note that a relation can be neither reflexive nor irreflexive. (Contributed by David A. Wheeler, 1-Dec-2019.)
⊢ (𝑅Irreflexive𝐴 ↔ (𝑅 ⊆ (𝐴 × 𝐴) ∧ ∀𝑥 ∈ 𝐴 ¬ 𝑥𝑅𝑥))
21.51.9  Algebra helpers This is an experimental approach to make it clearer (and easier) to do basic algebra in set.mm. These little theorems support basic algebra on equations at a slightly higher conceptual level. Instead of always having to "build up" equivalent expressions for one side of an equation, these theorems allow you to directly manipulate an equality. These higher-level steps lead to easier to understand proofs when they can be used, as well as proofs that are slightly shorter (when measured in steps). There are disadvantages. In particular, this approach requires many theorems (for many permutations to provide all of the operations). It can also only handle certain cases; more complex approaches must still be approached by "building up" equalities as is done today. However, I expect that we can create enough theorems to make it worth doing. I'm trying this out to see if this is helpful and if the number of permutations is manageable. To commute LHS for addition, use addcomli 11326. We might want to switch to a naming convention like addcomli 11326.
Theorem	mvlraddi 50204	Move the right term in a sum on the LHS to the RHS. (Contributed by David A. Wheeler, 11-Oct-2018.)
⊢ 𝐴 ∈ ℂ    &   ⊢ 𝐵 ∈ ℂ    &   ⊢ (𝐴 + 𝐵) = 𝐶    ⇒   ⊢ 𝐴 = (𝐶 − 𝐵)
Theorem	assraddsubi 50205	Associate RHS addition-subtraction. (Contributed by David A. Wheeler, 11-Oct-2018.)
⊢ 𝐵 ∈ ℂ    &   ⊢ 𝐶 ∈ ℂ    &   ⊢ 𝐷 ∈ ℂ    &   ⊢ 𝐴 = ((𝐵 + 𝐶) − 𝐷)    ⇒   ⊢ 𝐴 = (𝐵 + (𝐶 − 𝐷))
Theorem	joinlmuladdmuli 50206	Join AB+CB into (A+C) on LHS. (Contributed by David A. Wheeler, 26-Oct-2019.)
⊢ 𝐴 ∈ ℂ    &   ⊢ 𝐵 ∈ ℂ    &   ⊢ 𝐶 ∈ ℂ    &   ⊢ ((𝐴 · 𝐵) + (𝐶 · 𝐵)) = 𝐷    ⇒   ⊢ ((𝐴 + 𝐶) · 𝐵) = 𝐷
Theorem	joinlmulsubmuld 50207	Join AB-CB into (A-C) on LHS. (Contributed by David A. Wheeler, 15-Oct-2018.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → ((𝐴 · 𝐵) − (𝐶 · 𝐵)) = 𝐷)    ⇒   ⊢ (𝜑 → ((𝐴 − 𝐶) · 𝐵) = 𝐷)
Theorem	joinlmulsubmuli 50208	Join AB-CB into (A-C) on LHS. (Contributed by David A. Wheeler, 11-Oct-2018.)
⊢ 𝐴 ∈ ℂ    &   ⊢ 𝐵 ∈ ℂ    &   ⊢ 𝐶 ∈ ℂ    &   ⊢ ((𝐴 · 𝐵) − (𝐶 · 𝐵)) = 𝐷    ⇒   ⊢ ((𝐴 − 𝐶) · 𝐵) = 𝐷
Theorem	mvlrmuld 50209	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 50210	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)
21.51.10  Algebra helper examples Examples using the algebra helpers.
Theorem	i2linesi 50211	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 50212	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)))
21.51.11  Formal methods "surprises" 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 50213, empty-surprise 50215, and eximp-surprise 50217.
Theorem	alimp-surprise 50213	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 854). 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 50214) or using the allsome quantifier (df-alsi 50221) . For other "surprises" for new users of classical logic, see empty-surprise 50215 and eximp-surprise 50217. (Contributed by David A. Wheeler, 17-Oct-2018.)
⊢ ¬ ∃𝑥𝜑    ⇒   ⊢ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))
Theorem	alimp-no-surprise 50214	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 50213. The allsome quantifier also counters this problem, see df-alsi 50221. (Contributed by David A. Wheeler, 27-Oct-2018.)
⊢ ¬ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)
Theorem	empty-surprise 50215	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 3053). Thus, if 𝐴 is the empty set, this expression is always true regardless of the value of 𝜑 (see alimp-surprise 50213). 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 4294, 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/ 4294). 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 2670, celaront 2672, and cesaro 2679. For another "surprise" for new users of classical logic, see alimp-surprise 50213 and eximp-surprise 50217. (Contributed by David A. Wheeler, 20-Oct-2018.)
⊢ ¬ ∃𝑥 𝑥 ∈ 𝐴    ⇒   ⊢ ∀𝑥 ∈ 𝐴 𝜑
Theorem	empty-surprise2 50216	"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 50215. Here we prove an extreme example: we "prove" that false is true. Of course, we actually do no such thing (see notfal 1570); 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 50222. (Contributed by David A. Wheeler, 20-Oct-2018.)
⊢ ¬ ∃𝑥 𝑥 ∈ 𝐴    ⇒   ⊢ ∀𝑥 ∈ 𝐴 ⊥
Theorem	eximp-surprise 50217	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 854, 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 50218. See also alimp-surprise 50213 and empty-surprise 50215. (Contributed by David A. Wheeler, 17-Oct-2018.)
⊢ (∃𝑥(𝜑 → 𝜓) ↔ ∃𝑥(¬ 𝜑 ∨ 𝜓))
Theorem	eximp-surprise2 50218	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 50217, which shows what implication really expands to. See also empty-surprise 50215. (Contributed by David A. Wheeler, 18-Oct-2018.)
⊢ ∃𝑥 ¬ 𝜑    ⇒   ⊢ ∃𝑥(𝜑 → 𝜓)
21.51.12  Allsome quantifier 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 50213 and empty-surprise 50215 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 50215 I hope that others will eventually agree that allsome is awesome.
Syntax	walsi 50219	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 50220	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 50221	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 50222	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 50223	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 50224	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 50225	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 50226	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 50227	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 50228*	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 50229	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 50221; the proof itself builds on alimp-no-surprise 50214. For a contrast, see alimp-surprise 50213. (Contributed by David A. Wheeler, 27-Oct-2018.)
⊢ ¬ (∀!𝑥(𝜑 → 𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓))
21.51.13  Miscellaneous Miscellaneous proofs.
Theorem	5m4e1 50230	Prove that 5 - 4 = 1. (Contributed by David A. Wheeler, 31-Jan-2017.)
⊢ (5 − 4) = 1
Theorem	2p2ne5 50231	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 50232	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 50233	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)
21.51.14  Theorems about algebraic numbers
Theorem	aacllem 50234*	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)
21.52  Mathbox for Kunhao Zheng
21.52.1  Weighted AM-GM inequality
Theorem	amgmwlem 50235	Weighted version of amgmlem 26940. (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 50236	Alternate proof of amgmlem 26940 using amgmwlem 50235. (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 50237	Weighted arithmetic-geometric mean inequality for 𝑛 = 2 (compare amgm2d 44628). (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 50238	Young's inequality for 𝑛 = 2, a direct application of amgmw2d 50237. (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)))

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