P. 468 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 46701-46765   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	recotcl 46701	The closure of the cotangent function with a real argument. (Contributed by David A. Wheeler, 15-Mar-2014.)
⊢ ((𝐴 ∈ ℝ ∧ (sin‘𝐴) ≠ 0) → (cot‘𝐴) ∈ ℝ)
Theorem	recsec 46702	The reciprocal of secant is cosine. (Contributed by David A. Wheeler, 14-Mar-2014.)
⊢ ((𝐴 ∈ ℂ ∧ (cos‘𝐴) ≠ 0) → (cos‘𝐴) = (1 / (sec‘𝐴)))
Theorem	reccsc 46703	The reciprocal of cosecant is sine. (Contributed by David A. Wheeler, 14-Mar-2014.)
⊢ ((𝐴 ∈ ℂ ∧ (sin‘𝐴) ≠ 0) → (sin‘𝐴) = (1 / (csc‘𝐴)))
Theorem	reccot 46704	The reciprocal of cotangent is tangent. (Contributed by David A. Wheeler, 21-Mar-2014.)
⊢ ((𝐴 ∈ ℂ ∧ (sin‘𝐴) ≠ 0 ∧ (cos‘𝐴) ≠ 0) → (tan‘𝐴) = (1 / (cot‘𝐴)))
Theorem	rectan 46705	The reciprocal of tangent is cotangent. (Contributed by David A. Wheeler, 21-Mar-2014.)
⊢ ((𝐴 ∈ ℂ ∧ (sin‘𝐴) ≠ 0 ∧ (cos‘𝐴) ≠ 0) → (cot‘𝐴) = (1 / (tan‘𝐴)))
Theorem	sec0 46706	The value of the secant function at zero is one. (Contributed by David A. Wheeler, 16-Mar-2014.)
⊢ (sec‘0) = 1
Theorem	onetansqsecsq 46707	Prove the tangent squared secant squared identity (1 + ((tan A ) ^ 2 ) ) = ( ( sec 𝐴)↑2)). (Contributed by David A. Wheeler, 25-May-2015.)
⊢ ((𝐴 ∈ ℂ ∧ (cos‘𝐴) ≠ 0) → (1 + ((tan‘𝐴)↑2)) = ((sec‘𝐴)↑2))
Theorem	cotsqcscsq 46708	Prove the tangent squared cosecant squared identity (1 + ((cot A ) ^ 2 ) ) = ( ( csc 𝐴)↑2)). (Contributed by David A. Wheeler, 27-May-2015.)
⊢ ((𝐴 ∈ ℂ ∧ (sin‘𝐴) ≠ 0) → (1 + ((cot‘𝐴)↑2)) = ((csc‘𝐴)↑2))
20.44.5  Identities for "if" Utility theorems for "if".
Theorem	ifnmfalse 46709	If A is not a member of B, but an "if" condition requires it, then the "false" branch results. This is a simple utility to provide a slight shortening and simplification of proofs versus applying iffalse 4474 directly in this case. (Contributed by David A. Wheeler, 15-May-2015.)
⊢ (𝐴 ∉ 𝐵 → if(𝐴 ∈ 𝐵, 𝐶, 𝐷) = 𝐷)
20.44.6  Logarithms generalized to arbitrary base using ` logb ` Most of this subsection was moved to main set.mm, section "Logarithms to an arbitrary base".
Theorem	logb2aval 46710	Define the value of the logb function, the logarithm generalized to an arbitrary base, when used in the 2-argument form logb ⟨𝐵, 𝑋⟩ (Contributed by David A. Wheeler, 21-Jan-2017.) (Revised by David A. Wheeler, 16-Jul-2017.)
⊢ ((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ 𝑋 ∈ (ℂ ∖ {0})) → ( logb ‘⟨𝐵, 𝑋⟩) = ((log‘𝑋) / (log‘𝐵)))
20.44.7  Logarithm laws generalized to an arbitrary base - log_ Define "log using an arbitrary base" function and then prove some of its properties. This builds on previous work by Stefan O'Rear. This supports the notational form ((log_‘𝐵)‘𝑋); that looks a little more like traditional notation, but is different from other 2-parameter functions. E.g., ((log_‘;10)‘;;100) = 2. This form is less convenient to work with inside set.mm as compared to the (𝐵 logb 𝑋) form defined separately.
Syntax	clog- 46711	Extend class notation to include the logarithm generalized to an arbitrary base.
class log_
Definition	df-logbALT 46712*	Define the log_ operator. This is the logarithm generalized to an arbitrary base. It can be used as ((log_‘𝐵)‘𝑋) for "log base B of X". This formulation suggested by Mario Carneiro. (Contributed by David A. Wheeler, 14-Jul-2017.) (New usage is discouraged.)
⊢ log_ = (𝑏 ∈ (ℂ ∖ {0, 1}) ↦ (𝑥 ∈ (ℂ ∖ {0}) ↦ ((log‘𝑥) / (log‘𝑏))))
20.44.8  Formally define notions such as reflexivity EXPERIMENTAL. Several terms are used in comments but not directly defined in set.mm. For example, there are proofs that a number of specific relations are reflexive, but there is no formal definition of what being reflexive actually *means*. Stating the relationships directly, instead of defining a broader property such as being reflexive, can reduce proof size (because the definition of that property does not need to be expanded later). A disadvantage, however, is that there are several terms that are widely used in comments but do not have a clear formal definition. Here we define wffs that formally define some of these key terms. The intent isn't to use these directly, but to instead provide a clear formal definition of widely-used mathematical terminology (we even use this terminology within the comments of set.mm itself). We could define these using extensible structures, but doing so appears overly restrictive. These definitions don't require the use of extensible structures; requiring something to be in an extensible structure to use them is too restrictive. Even if an extensible structure is already in use, it may in use for other things. For example, in geometry, there is a "less-than" relation, but while the geometry itself is an extensible structure, we would have to build a new structure to state "the geometric less-than relation is transitive" (which is more work than it's probably worth). By creating definitions that aren't tied to extensible structures we create definitions that can be applied to anything, including extensible structures, in whatever way we'd like. BJ suggests that it might be better to define these as functions. There are many advantages to doing that, but they won't work for proper classes. I'm currently trying to also support proper classes, so I have not taken that approach, but if that turns out to be unreasonable then BJ's approach is very much worth considering. Examples would be: BinRel = (𝑥 ∈ V ↦ {𝑟 ∣ 𝑟 ⊆ (𝑥 × 𝑥)}), ReflBinRel = (𝑥 ∈ V ↦ {𝑟 ∈ ( BinRel ‘𝑥) ∣ ( I ↾ 𝑥) ⊆ 𝑟}), and IrreflBinRel = (𝑥 ∈ V ↦ {𝑟 ∈ ( BinRel ‘𝑥) ∣ (𝑟 ∩ ( I ↾ 𝑥)) = ∅}). For more discussion see: https://github.com/metamath/set.mm/pull/1286
Syntax	wreflexive 46713	Extend wff definition to include "Reflexive" applied to a class, which is true iff class R is a reflexive relation over the set A. See df-reflexive 46714. (Contributed by David A. Wheeler, 1-Dec-2019.)
wff 𝑅Reflexive𝐴
Definition	df-reflexive 46714*	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 46715	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 46716. (Contributed by David A. Wheeler, 1-Dec-2019.)
wff 𝑅Irreflexive𝐴
Definition	df-irreflexive 46716*	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))
20.44.9  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 11217. We might want to switch to a naming convention like addcomli 11217.
Theorem	comraddi 46717	Commute RHS addition. See addcomli 11217 to commute addition on LHS. (Contributed by David A. Wheeler, 11-Oct-2018.)
⊢ 𝐵 ∈ ℂ    &   ⊢ 𝐶 ∈ ℂ    &   ⊢ 𝐴 = (𝐵 + 𝐶)    ⇒   ⊢ 𝐴 = (𝐶 + 𝐵)
Theorem	mvlraddi 46718	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	mvrladdi 46719	Move the left term in a sum on the RHS to the LHS. (Contributed by David A. Wheeler, 11-Oct-2018.)
⊢ 𝐵 ∈ ℂ    &   ⊢ 𝐶 ∈ ℂ    &   ⊢ 𝐴 = (𝐵 + 𝐶)    ⇒   ⊢ (𝐴 − 𝐵) = 𝐶
Theorem	assraddsubi 46720	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 46721	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 46722	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 46723	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 46724	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 46725	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)
20.44.10  Algebra helper examples Examples using the algebra helpers.
Theorem	i2linesi 46726	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 46727	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)))
20.44.11  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 46728, empty-surprise 46730, and eximp-surprise 46732.
Theorem	alimp-surprise 46728	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 851). 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 46729) or using the allsome quantifier (df-alsi 46736) . For other "surprises" for new users of classical logic, see empty-surprise 46730 and eximp-surprise 46732. (Contributed by David A. Wheeler, 17-Oct-2018.)
⊢ ¬ ∃𝑥𝜑    ⇒   ⊢ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))
Theorem	alimp-no-surprise 46729	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 46728. The allsome quantifier also counters this problem, see df-alsi 46736. (Contributed by David A. Wheeler, 27-Oct-2018.)
⊢ ¬ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)
Theorem	empty-surprise 46730	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 3063). Thus, if 𝐴 is the empty set, this expression is always true regardless of the value of 𝜑 (see alimp-surprise 46728). 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 4286, 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/ 4286). 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 2668, celaront 2670, and cesaro 2677. For another "surprise" for new users of classical logic, see alimp-surprise 46728 and eximp-surprise 46732. (Contributed by David A. Wheeler, 20-Oct-2018.)
⊢ ¬ ∃𝑥 𝑥 ∈ 𝐴    ⇒   ⊢ ∀𝑥 ∈ 𝐴 𝜑
Theorem	empty-surprise2 46731	"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 46730. Here we prove an extreme example: we "prove" that false is true. Of course, we actually do no such thing (see notfal 1567); 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 46737. (Contributed by David A. Wheeler, 20-Oct-2018.)
⊢ ¬ ∃𝑥 𝑥 ∈ 𝐴    ⇒   ⊢ ∀𝑥 ∈ 𝐴 ⊥
Theorem	eximp-surprise 46732	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 851, 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 46733. See also alimp-surprise 46728 and empty-surprise 46730. (Contributed by David A. Wheeler, 17-Oct-2018.)
⊢ (∃𝑥(𝜑 → 𝜓) ↔ ∃𝑥(¬ 𝜑 ∨ 𝜓))
Theorem	eximp-surprise2 46733	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 46732, which shows what implication really expands to. See also empty-surprise 46730. (Contributed by David A. Wheeler, 18-Oct-2018.)
⊢ ∃𝑥 ¬ 𝜑    ⇒   ⊢ ∃𝑥(𝜑 → 𝜓)
20.44.12  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 46728 and empty-surprise 46730 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 46730 I hope that others will eventually agree that allsome is awesome.
Syntax	walsi 46734	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 46735	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 46736	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 46737	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 46738	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 46739	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 46740	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 46741	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 46742	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 46743*	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 46744	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 46736; the proof itself builds on alimp-no-surprise 46729. For a contrast, see alimp-surprise 46728. (Contributed by David A. Wheeler, 27-Oct-2018.)
⊢ ¬ (∀!𝑥(𝜑 → 𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓))
20.44.13  Miscellaneous Miscellaneous proofs.
Theorem	5m4e1 46745	Prove that 5 - 4 = 1. (Contributed by David A. Wheeler, 31-Jan-2017.)
⊢ (5 − 4) = 1
Theorem	2p2ne5 46746	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 46747	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 46748	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)
20.44.14  Theorems about algebraic numbers
Theorem	aacllem 46749*	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)
20.45  Mathbox for Kunhao Zheng
20.45.1  Weighted AM-GM inequality
Theorem	amgmwlem 46750	Weighted version of amgmlem 26188. (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 46751	Alternate proof of amgmlem 26188 using amgmwlem 46750. (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 46752	Weighted arithmetic-geometric mean inequality for 𝑛 = 2 (compare amgm2d 42022). (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 46753	Young's inequality for 𝑛 = 2, a direct application of amgmw2d 46752. (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)))
20.46  Mathbox for Ender Ting
20.46.1  Increasing sequences and subsequences
Theorem	et-ltneverrefl 46754	Less-than class is never reflexive. (Contributed by Ender Ting, 22-Nov-2024.) Prefer to specify theorem domain and then apply ltnri 11134. (New usage is discouraged.)
⊢ ¬ 𝐴 < 𝐴
Theorem	natlocalincr 46755*	Global monotonicity on half-open range implies local monotonicity. (Contributed by Ender Ting, 22-Nov-2024.)
⊢ ∀𝑘 ∈ (0..^𝑇)∀𝑡 ∈ (1..^(𝑇 + 1))(𝑘 < 𝑡 → (𝐵‘𝑘) < (𝐵‘𝑡))    ⇒   ⊢ ∀𝑘 ∈ (0..^𝑇)(𝐵‘𝑘) < (𝐵‘(𝑘 + 1))
Theorem	natglobalincr 46756*	Local monotonicity on half-open integer range implies global monotonicity. (Contributed by Ender Ting, 23-Nov-2024.)
⊢ ∀𝑘 ∈ (0..^𝑇)(𝐵‘𝑘) < (𝐵‘(𝑘 + 1))    &   ⊢ 𝑇 ∈ ℤ    ⇒   ⊢ ∀𝑘 ∈ (0..^𝑇)∀𝑡 ∈ ((𝑘 + 1)...𝑇)(𝐵‘𝑘) < (𝐵‘𝑡)
Syntax	cupword 46757	Extend class notation to include the set of strictly increasing sequences.
class UpWord𝑆
Definition	df-upword 46758*	Strictly increasing sequence is a sequence, adjacent elements of which increase. (Contributed by Ender Ting, 19-Nov-2024.)
⊢ UpWord𝑆 = {𝑤 ∣ (𝑤 ∈ Word 𝑆 ∧ ∀𝑘 ∈ (0..^((♯‘𝑤) − 1))(𝑤‘𝑘) < (𝑤‘(𝑘 + 1)))}
Theorem	upwordnul 46759	Empty set is an increasing sequence for every range. (Contributed by Ender Ting, 19-Nov-2024.)
⊢ ∅ ∈ UpWord𝑆
Theorem	upwordisword 46760	Any increasing sequence is a sequence. (Contributed by Ender Ting, 19-Nov-2024.)
⊢ (𝐴 ∈ UpWord𝑆 → 𝐴 ∈ Word 𝑆)
Theorem	singoutnword 46761	Singleton with character out of range 𝑉 is not a word for that range. (Contributed by Ender Ting, 21-Nov-2024.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (¬ 𝐴 ∈ 𝑉 → ¬ ⟨“𝐴”⟩ ∈ Word 𝑉)
Theorem	singoutnupword 46762	Singleton with character out of range 𝑆 is not an increasing sequence for that range. (Contributed by Ender Ting, 22-Nov-2024.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (¬ 𝐴 ∈ 𝑆 → ¬ ⟨“𝐴”⟩ ∈ UpWord𝑆)
Theorem	upwordsing 46763	Singleton is an increasing sequence for any compatible range. (Contributed by Ender Ting, 21-Nov-2024.)
⊢ 𝐴 ∈ 𝑆    ⇒   ⊢ ⟨“𝐴”⟩ ∈ UpWord𝑆
Theorem	upwordsseti 46764	Strictly increasing sequences with a set given for range form a set. (Contributed by Ender Ting, 21-Nov-2024.)
⊢ 𝑆 ∈ V    ⇒   ⊢ UpWord𝑆 ∈ V
Theorem	tworepnotupword 46765	Word of two matching characters is never an increasing sequence. (Contributed by Ender Ting, 22-Nov-2024.)
⊢ 𝐴 ∈ V    ⇒   ⊢ ¬ (⟨“𝐴”⟩ ++ ⟨“𝐴”⟩) ∈ UpWord𝑆

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