P. 508 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 50701-50800   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	assraddsubi 50701	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 50702	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 50703	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 50704	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 50705	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 50706	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)
21.53.10  Algebra helper examples Examples using the algebra helpers.
Theorem	i2linesi 50707	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 50708	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)))
21.53.11  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 50709, empty-surprise 50711, and eximp-surprise 50713.
Theorem	alimp-surprise 50709	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 867). 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 50710) or using the allsome quantifier (df-als 50717) . For other "surprises" for new users of classical logic, see empty-surprise 50711 and eximp-surprise 50713. (Contributed by David A. Wheeler, 17-Oct-2018.)
⊢ ¬ ∃𝑥𝜑    ⇒   ⊢ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))
Theorem	alimp-no-surprise 50710	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 50709. The allsome quantifier also counters this problem, see df-als 50717. (Contributed by David A. Wheeler, 27-Oct-2018.)
⊢ ¬ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)
Theorem	empty-surprise 50711	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 3077). Thus, if 𝐴 is the empty set, this expression is always true regardless of the value of 𝜑 (see alimp-surprise 50709). 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 4300, 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/ 4300). 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 2693, celaront 2695, and cesaro 2702. For another "surprise" for new users of classical logic, see alimp-surprise 50709 and eximp-surprise 50713. (Contributed by David A. Wheeler, 20-Oct-2018.)
⊢ ¬ ∃𝑥 𝑥 ∈ 𝐴    ⇒   ⊢ ∀𝑥 ∈ 𝐴 𝜑
Theorem	empty-surprise2 50712	"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 50711. Here we prove an extreme example: we "prove" that false is true. Of course, we actually do no such thing (see notfal 1598); 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-rals 50718. (Contributed by David A. Wheeler, 20-Oct-2018.)
⊢ ¬ ∃𝑥 𝑥 ∈ 𝐴    ⇒   ⊢ ∀𝑥 ∈ 𝐴 ⊥
Theorem	eximp-surprise 50713	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 867, 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 50714. See also alimp-surprise 50709 and empty-surprise 50711. (Contributed by David A. Wheeler, 17-Oct-2018.)
⊢ (∃𝑥(𝜑 → 𝜓) ↔ ∃𝑥(¬ 𝜑 ∨ 𝜓))
Theorem	eximp-surprise2 50714	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 50713, which shows what implication really expands to. See also empty-surprise 50711. (Contributed by David A. Wheeler, 18-Oct-2018.)
⊢ ∃𝑥 ¬ 𝜑    ⇒   ⊢ ∃𝑥(𝜑 → 𝜓)
21.53.12  Allsome quantifier These are definitions and proofs involving the "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 50709 and empty-surprise 50711 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 ∈ 𝐴(𝜑 → 𝜓). 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. Naming: "als" is allsome. The form restricted to a class is prefixed with "r", following the way set.mm names the restricted quantifiers it is built from: ∀ gives df-ral 3077 and ∃ gives df-rex 3087, so df-als 50717 (the general form) gives df-rals 50718 (the restricted form). Earlier versions of this material differed, so old references may not match. They wrote the quantifier as an "inverted A" followed by an exclamation point, and they named the general form df-alsi and the restricted form df-alsc. The symbol is now an "inverted A" followed by a "backwards E", which more readers can correctly guess without being taught it. The restricted definition also changed, and the older one was a mistake; see df-rals 50718 for what was wrong with it. Soundness of the two definitions below. Definitions are required to be eliminable and conservative (see the section comment for df-bi 210). Both df-als 50717 and df-rals 50718 meet these requirements directly, and so neither needs a justification theorem. Each is stated as a biconditional whose left side is a new syntax construct (wals 50715 or wrals 50716) applied to distinct metavariables, and whose right side uses only constructs introduced earlier (∀, ∃, ∧, →, and the restricted quantifiers df-ral 3077 and df-rex 3087). Any occurrence of the new construct can therefore be replaced by the right side, which is eliminability; conservativity follows, since a proof of a statement not mentioning ∀∃ can have every use of the definition replaced in this way. Note in particular that every variable occurring on the right side already occurs on the left side, so no dummy variable is introduced. A justification theorem is needed only when that fails, that is, when a definition introduces a dummy variable and the choice of that variable must be shown not to matter (as in eujust 2596 for df-eu 2594), or when the definition cannot use ↔ because it is defining ↔ itself (as in bijust 208 for df-bi 210). Neither case arises here. The restricted quantifier definitions df-ral 3077 and df-rex 3087 have the same shape as these and likewise need no justification. For more, see "The Allsome Quantifier" by David A. Wheeler at https://dwheeler.com/essays/allsome.html 3087 I hope that others will eventually agree that allsome is awesome.
Syntax	wals 50715	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 one 𝑥 where 𝜑 is true. (Contributed by David A. Wheeler, 20-Oct-2018.) (Revised by David A. Wheeler, 12-Jul-2026.)
wff ∀∃𝑥(𝜑 → 𝜓)
Syntax	wrals 50716	Extend wff definition to include "all some" applied to a class, which means 𝜓 is true whenever 𝜑 is true for 𝑥 in 𝐴, and there is at least one 𝑥 in 𝐴 where 𝜑 is true. (Contributed by David A. Wheeler, 20-Oct-2018.) (Revised by David A. Wheeler, 12-Jul-2026.)
wff ∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓)
Definition	df-als 50717	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-rals 50718	Define "all some" applied to a class, which means 𝜓 is true whenever 𝜑 is true for 𝑥 in 𝐴, and there is at least one 𝑥 in 𝐴 where 𝜑 is true. An older definition of the "all some" quantifier when scoped to a class, named df-alsc and now removed, instead applied a bare formula 𝜑 to the members of a class, asserting only (∀𝑥 ∈ 𝐴𝜑 ∧ ∃𝑥𝑥 ∈ 𝐴), that is, that the formula held throughout 𝐴 and that 𝐴 had at least one member. I've now decided that that was a mistake. Its older existence conjunct ∃𝑥𝑥 ∈ 𝐴 did not require any member of 𝐴 to satisfy the antecedent, so if the formula was itself an implication, that inner implication could still be vacuously true, which is precisely what the allsome quantifier exists to prevent. For example, the older definition meant that "among Martians, all tall ones are green" could be considered true if there are Martians, but no tall Martians. This version of the definition instead ensures that claims of the form "among Martians, all tall ones are green" can only be true if all tall Martians are green and that there is at least one tall Martian. (Contributed by David A. Wheeler, 20-Oct-2018.) (Revised by David A. Wheeler, 12-Jul-2026.)
⊢ (∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ (∀𝑥 ∈ 𝐴 (𝜑 → 𝜓) ∧ ∃𝑥 ∈ 𝐴 𝜑))
Theorem	dfrals2 50719	The bounded "all some" form is the general form with the class membership folded into the antecedent. (Contributed by David A. Wheeler, 22-Oct-2018.) (Revised by David A. Wheeler, 12-Jul-2026.)
⊢ (∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ ∀∃𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) → 𝜓))
Theorem	alsd 50720	Introduction rule: "all some" holds if the "for all" part holds and the antecedent has a witness. This is the converse of als1d 50722 and als2d 50723 taken together, and is what lets an "all some" statement be proved rather than merely taken apart. (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ (𝜑 → ∀𝑥(𝜓 → 𝜒))    &   ⊢ (𝜑 → ∃𝑥𝜓)    ⇒   ⊢ (𝜑 → ∀∃𝑥(𝜓 → 𝜒))
Theorem	ralsd 50721	Introduction rule for "all some" restricted to a class. This is the converse of rals1d 50724 and rals2d 50725 taken together. (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝜓 → 𝜒))    &   ⊢ (𝜑 → ∃𝑥 ∈ 𝐴 𝜓)    ⇒   ⊢ (𝜑 → ∀∃𝑥 ∈ 𝐴(𝜓 → 𝜒))
Theorem	als1d 50722	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	als2d 50723	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	rals1d 50724	Deduction rule: Given "all some" applied to a class, you can extract the "for all" part. (Contributed by David A. Wheeler, 20-Oct-2018.) (Revised by David A. Wheeler, 12-Jul-2026.)
⊢ (𝜑 → ∀∃𝑥 ∈ 𝐴(𝜓 → 𝜒))    ⇒   ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝜓 → 𝜒))
Theorem	rals2d 50725	Deduction rule: Given "all some" applied to a class, you can extract the "there exists" part. Note that the witness must satisfy the antecedent 𝜓, not merely be a member of 𝐴. (Contributed by David A. Wheeler, 20-Oct-2018.) (Revised by David A. Wheeler, 12-Jul-2026.)
⊢ (𝜑 → ∀∃𝑥 ∈ 𝐴(𝜓 → 𝜒))    ⇒   ⊢ (𝜑 → ∃𝑥 ∈ 𝐴 𝜓)
Theorem	ralsn0d 50726*	Deduction rule: Given "all some" applied to a class, the class is not the empty set. (Contributed by David A. Wheeler, 23-Oct-2018.) (Revised by David A. Wheeler, 12-Jul-2026.)
⊢ (𝜑 → ∀∃𝑥 ∈ 𝐴(𝜓 → 𝜒))    ⇒   ⊢ (𝜑 → 𝐴 ≠ ∅)
Theorem	alsex 50727	The consequent of an "all some" is witnessed: if 𝜓 holds of every 𝑥 satisfying 𝜑, and some 𝑥 satisfies 𝜑, then some 𝑥 satisfies 𝜓. This is the positive counterpart of als-no-surprise 50735, and it is the property that ordinary "for all" with implication lacks: from ∀𝑥(𝜑 → 𝜓) alone nothing whatever follows about 𝜓, as alimp-surprise 50709 shows. It is the reason the allsome quantifier says what a speaker of "all Martians are green" usually means. (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ (∀∃𝑥(𝜑 → 𝜓) → ∃𝑥𝜓)
Theorem	ralsex 50728	The consequent of an "all some" restricted to a class is witnessed: some member of 𝐴 satisfying 𝜑 also satisfies 𝜓. Restricted counterpart of alsex 50727. (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ (∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) → ∃𝑥 ∈ 𝐴 𝜓)
Theorem	alsbii 50729	Congruence: equivalents may be substituted inside an "all some". (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ (𝜑 ↔ 𝜒)    &   ⊢ (𝜓 ↔ 𝜃)    ⇒   ⊢ (∀∃𝑥(𝜑 → 𝜓) ↔ ∀∃𝑥(𝜒 → 𝜃))
Theorem	ralsbii 50730	Congruence for "all some" restricted to a class. (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ (𝜑 ↔ 𝜒)    &   ⊢ (𝜓 ↔ 𝜃)    ⇒   ⊢ (∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ ∀∃𝑥 ∈ 𝐴(𝜒 → 𝜃))
Theorem	alsbid 50731	Deduction form of alsbii 50729. (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ Ⅎ𝑥𝜑    &   ⊢ (𝜑 → (𝜓 ↔ 𝜃))    &   ⊢ (𝜑 → (𝜒 ↔ 𝜏))    ⇒   ⊢ (𝜑 → (∀∃𝑥(𝜓 → 𝜒) ↔ ∀∃𝑥(𝜃 → 𝜏)))
Theorem	nfals 50732	Bound-variable hypothesis builder for "all some". (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ Ⅎ𝑥𝜑    &   ⊢ Ⅎ𝑥𝜓    ⇒   ⊢ Ⅎ𝑥∀∃𝑦(𝜑 → 𝜓)
Theorem	nfrals 50733*	Bound-variable hypothesis builder for "all some" restricted to a class. (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ Ⅎ𝑥𝐴    &   ⊢ Ⅎ𝑥𝜑    &   ⊢ Ⅎ𝑥𝜓    ⇒   ⊢ Ⅎ𝑥∀∃𝑦 ∈ 𝐴(𝜑 → 𝜓)
Theorem	cbvals 50734*	Rule used to change bound variables, using implicit substitution. (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜒))    &   ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜃))    ⇒   ⊢ (∀∃𝑥(𝜑 → 𝜓) ↔ ∀∃𝑦(𝜒 → 𝜃))
Theorem	als-no-surprise 50735	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-als 50717; the proof itself builds on alimp-no-surprise 50710. For a contrast, see alimp-surprise 50709. (Contributed by David A. Wheeler, 27-Oct-2018.)
⊢ ¬ (∀∃𝑥(𝜑 → 𝜓) ∧ ∀∃𝑥(𝜑 → ¬ 𝜓))
Theorem	rals-no-surprise 50736	Demonstrate that there is never a "surprise" when using the allsome quantifier restricted to a class, that is, it is never possible for the consequent to be both always true and always false of the members of 𝐴 that satisfy the antecedent. This is the restricted counterpart of als-no-surprise 50735, and follows from it by dfrals2 50719. Note that this needs no assumption that 𝐴 is nonempty, because allsome requires a member of 𝐴 satisfying 𝜑, and that member would have to satisfy both 𝜓 and ¬ 𝜓. The ordinary restricted "for all" requires no such member and can be vacuously true, as shown in empty-surprise2 50712; that is the point of allsome. (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ ¬ (∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ∧ ∀∃𝑥 ∈ 𝐴(𝜑 → ¬ 𝜓))
Theorem	ralrals 50737	If the universal part of a restricted "all some" statement holds, then the statement reduces to the existence of a member of 𝐴 satisfying its antecedent. This is the restricted counterpart of ralals 50743. (Contributed by Peter Mazsa and David A. Wheeler, 20-Jul-2026.)
⊢ (∀𝑥 ∈ 𝐴 (𝜑 → 𝜓) → (∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ ∃𝑥 ∈ 𝐴 𝜑))
Theorem	rexrals 50738	If a member of 𝐴 satisfying the antecedent exists, then a restricted "all some" statement reduces to its universal part. This is the restricted counterpart of rexals 50744. (Contributed by Peter Mazsa and David A. Wheeler, 20-Jul-2026.)
⊢ (∃𝑥 ∈ 𝐴 𝜑 → (∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ ∀𝑥 ∈ 𝐴 (𝜑 → 𝜓)))
Theorem	alsanmo 50739	An "all some" statement conjoined with the claim that at most one 𝑥 satisfies its antecedent is equivalent to the universal part conjoined with the claim that exactly one 𝑥 satisfies the antecedent. The "all some" quantifier supplies the existence of such an 𝑥 and ∃*𝑥𝜑 supplies the at-most-one part, so together they yield ∃!𝑥𝜑. (Contributed by Peter Mazsa and David A. Wheeler, 20-Jul-2026.)
⊢ ((∀∃𝑥(𝜑 → 𝜓) ∧ ∃*𝑥𝜑) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃!𝑥𝜑))
Theorem	ralsanmo 50740	An "all some" statement restricted to a class, conjoined with the claim that at most one 𝑥 in 𝐴 satisfies its antecedent, is equivalent to the universal part conjoined with the claim that exactly one 𝑥 in 𝐴 satisfies the antecedent. This is the restricted counterpart of alsanmo 50739. (Contributed by Peter Mazsa and David A. Wheeler, 20-Jul-2026.)
⊢ ((∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ∧ ∃*𝑥 ∈ 𝐴 𝜑) ↔ (∀𝑥 ∈ 𝐴 (𝜑 → 𝜓) ∧ ∃!𝑥 ∈ 𝐴 𝜑))
Theorem	alsralrex 50741*	The general "all some" quantifier with class membership as its antecedent holds if and only if 𝜑 holds for every 𝑥 in 𝐴 and some 𝑥 in 𝐴 satisfies 𝜑. (Contributed by Peter Mazsa, 27-Nov-2018.) (Revised by David A. Wheeler, 15-Jul-2026.)
⊢ (∀∃𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑥 ∈ 𝐴 𝜑))
Theorem	alsraln0 50742*	The general "all some" quantifier with class membership as its antecedent holds if and only if 𝜑 holds for every 𝑥 in 𝐴 and 𝐴 is not empty. (Contributed by Peter Mazsa, 28-Nov-2018.) (Revised by David A. Wheeler, 15-Jul-2026.)
⊢ (∀∃𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ 𝐴 ≠ ∅))
Theorem	ralals 50743*	If 𝜑 holds for every 𝑥 in 𝐴, then the general "all some" quantifier with class membership as its antecedent reduces to the assertion that some 𝑥 in 𝐴 satisfies 𝜑. See ralrals 50737 for the restricted counterpart. (Contributed by Peter Mazsa, 19-Dec-2018.) (Revised by David A. Wheeler, 15-Jul-2026.)
⊢ (∀𝑥 ∈ 𝐴 𝜑 → (∀∃𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∃𝑥 ∈ 𝐴 𝜑))
Theorem	rexals 50744*	If some 𝑥 in 𝐴 satisfies 𝜑, then the general "all some" quantifier with class membership as its antecedent reduces to the assertion that 𝜑 holds for every 𝑥 in 𝐴. See rexrals 50738 for the restricted counterpart. (Contributed by Peter Mazsa, 19-Dec-2018.) (Revised by David A. Wheeler, 15-Jul-2026.)
⊢ (∃𝑥 ∈ 𝐴 𝜑 → (∀∃𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀𝑥 ∈ 𝐴 𝜑))
Theorem	n0als 50745*	If 𝐴 is not empty, then the general "all some" quantifier with class membership as its antecedent reduces to the assertion that 𝜑 holds for every 𝑥 in 𝐴. (Contributed by Peter Mazsa, 19-Dec-2018.) (Revised by David A. Wheeler, 15-Jul-2026.)
⊢ (𝐴 ≠ ∅ → (∀∃𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀𝑥 ∈ 𝐴 𝜑))
Theorem	2alsraln0 50746*	Nested general "all some" quantifiers with class membership as their antecedents: 𝜑 holds for every 𝑥 in 𝐴 and every 𝑦 in 𝐵, and both 𝐴 and 𝐵 are not empty. (Contributed by Peter Mazsa, 28-May-2019.) (Revised by David A. Wheeler, 15-Jul-2026.)
⊢ (∀∃𝑥(𝑥 ∈ 𝐴 → ∀∃𝑦(𝑦 ∈ 𝐵 → 𝜑)) ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 𝜑 ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)))
Theorem	2alsraln0id 50747*	Nested general "all some" quantifiers with class membership as their antecedents, for the same class 𝐴: 𝜑 holds for every 𝑥 and every 𝑦 in 𝐴, and 𝐴 is not empty. (Contributed by Peter Mazsa, 28-May-2019.) (Revised by David A. Wheeler, 15-Jul-2026.)
⊢ (∀∃𝑥(𝑥 ∈ 𝐴 → ∀∃𝑦(𝑦 ∈ 𝐴 → 𝜑)) ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 𝜑 ∧ 𝐴 ≠ ∅))
21.53.13  Allsome one quantifier These are definitions and proofs involving the "allsome one" quantifier, which extends the "allsome" quantifier of the previous section in the same way that ∃! (df-eu 2594) extends ∃. Some systems extend "there exists" by appending a character to it. If a system provides such an extension, it should provide it for allsome as well: append the same character, let it modify allsome's existence conjunct, and change nothing else. Appending "!" gives "allsome one", so ∀∃!𝑥(𝜑 → 𝜓) means that 𝜓 is true whenever 𝜑 is true and that exactly one 𝑥 satisfies 𝜑. This is what the English word "the" usually does. "The king is hungry" claims that a king exists, that there is only one, and that he is hungry, and the form ∀∃!𝑥(𝜑 → 𝜓) claims exactly that kind of statement (specifically when 𝜑 means 𝑥 is a king and 𝜓 means 𝑥 is hungry). English says all of that in a single phrase, and the first two have a dedicated word ("the") for the construct. Many other languages have their own dedicated way of saying this. Languages reserve that kind of compression for what their speakers need constantly, which is a good reason for a formal notation to be able to say it just as directly, rather than spelling it out afresh as a conjunction of two quantified formulas every time it comes up. Russell analyzed such definite descriptions with this apparatus in "On Denoting", where his example of a phrase that appears to refer to someone but in fact denotes no one at all is "the present King of France", item (1) of [Russell1905] p. 479, France being a republic that has no king. Write "the king is hungry" as ∀𝑥(𝜑 → 𝜓) instead and only the last of those three claims survives. The existence claim is silently gone, since that formula is vacuously true when there is no king, and the uniqueness claim is silently gone as well, since that formula holds just the same when there are five kings. Russell reached the opposite verdict for the same example, remarking that every proposition of the form "the present King of France ..." is false, [Russell1905] p. 482. The reason to care is the reason that motivates allsome, only more so; see alimp-surprise 50709 and empty-surprise 50711. Note that this is not merely a way of writing ∃!𝑥(𝜑 ∧ 𝜓). Reading 𝜑 as "is a king" and 𝜓 as "is hungry", ∃!𝑥(𝜑 ∧ 𝜓) says that there is exactly one hungry king, whereas ∀∃!𝑥(𝜑 → 𝜓) says that there is exactly one king and that he is hungry. The first is true in a region with five kings exactly one of whom is hungry; the second is false there. Uniqueness attaches to the antecedent, not to the conjunction. See dfalseu2 50765 for the exact relationship between the two and alseueu 50766 for the one direction that does hold. Naming: "alseu" is allsome ("als", as in df-als 50717) extended with "exactly one" ("eu", as in df-eu 2594). The form restricted to a class is prefixed with "r", following df-rals 50718 and df-reu 3366, giving df-ralseu 50751. Soundness: df-alseu 50750 and df-ralseu 50751 are eliminable and conservative directly, so neither needs a justification theorem. Definitions are required to be eliminable and conservative; see the section comment for df-bi 210. Each is a biconditional whose left side is a new syntax construct (walseu 50748 or wralseu 50749) applied to distinct metavariables, and whose right side uses only constructs introduced earlier (∀, ∃!, ∧, →, and the restricted quantifiers df-ral 3077 and df-reu 3366), so any occurrence of the new construct can be replaced by the right side, which is eliminability. Conservativity follows, since a proof of a statement not mentioning ∀∃! can have every use of the definition replaced in this way. Every variable occurring on the right side already occurs on the left side, so no dummy variable is introduced, and introducing a dummy variable whose choice must be shown not to matter is the only circumstance here that would call for a justification theorem. For more, see "The Allsome Quantifier" by David A. Wheeler at https://dwheeler.com/essays/allsome.html 3366
Syntax	walseu 50748	Extend wff definition to include "all some one" applied to a top-level implication, which means 𝜓 is true whenever 𝜑 is true, and exactly one 𝑥 satisfies 𝜑. (Contributed by David A. Wheeler, 21-Jul-2026.)
wff ∀∃!𝑥(𝜑 → 𝜓)
Syntax	wralseu 50749	Extend wff definition to include "all some one" applied to a class, which means 𝜓 is true whenever 𝜑 is true for 𝑥 in 𝐴, and exactly one 𝑥 in 𝐴 satisfies 𝜑. (Contributed by David A. Wheeler, 21-Jul-2026.)
wff ∀∃!𝑥 ∈ 𝐴(𝜑 → 𝜓)
Definition	df-alseu 50750	Define "all some one" applied to a top-level implication, which means 𝜓 is true whenever 𝜑 is true and exactly one 𝑥 satisfies 𝜑. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (∀∃!𝑥(𝜑 → 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃!𝑥𝜑))
Definition	df-ralseu 50751	Define "all some one" applied to a class, which means 𝜓 is true whenever 𝜑 is true for 𝑥 in 𝐴, and exactly one 𝑥 in 𝐴 satisfies 𝜑. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (∀∃!𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ (∀𝑥 ∈ 𝐴 (𝜑 → 𝜓) ∧ ∃!𝑥 ∈ 𝐴 𝜑))
Theorem	dfralseu2 50752	The bounded "all some one" form is the general form with the class membership folded into the antecedent. This is the "all some one" counterpart of dfrals2 50719. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (∀∃!𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ ∀∃!𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) → 𝜓))
Theorem	alseuals 50753	"All some one" implies "all some": requiring exactly one witness is stronger than requiring at least one. Any consequence of an allsome statement is therefore a consequence of the corresponding "all some one" statement, which is how alseu-no-surprise 50767 is proved. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (∀∃!𝑥(𝜑 → 𝜓) → ∀∃𝑥(𝜑 → 𝜓))
Theorem	ralseurals 50754	"All some one" restricted to a class implies "all some" restricted to that class. Restricted counterpart of alseuals 50753. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (∀∃!𝑥 ∈ 𝐴(𝜑 → 𝜓) → ∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓))
Theorem	alseud 50755	Introduction rule: "all some one" holds if the "for all" part holds and the antecedent has exactly one witness. This is the converse of alseu1d 50757 and alseu2d 50758 taken together. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (𝜑 → ∀𝑥(𝜓 → 𝜒))    &   ⊢ (𝜑 → ∃!𝑥𝜓)    ⇒   ⊢ (𝜑 → ∀∃!𝑥(𝜓 → 𝜒))
Theorem	ralseud 50756	Introduction rule for "all some one" restricted to a class. This is the converse of ralseu1d 50759 and ralseu2d 50760 taken together. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝜓 → 𝜒))    &   ⊢ (𝜑 → ∃!𝑥 ∈ 𝐴 𝜓)    ⇒   ⊢ (𝜑 → ∀∃!𝑥 ∈ 𝐴(𝜓 → 𝜒))
Theorem	alseu1d 50757	Deduction rule: Given "all some one" applied to a top-level inference, you can extract the "for all" part. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (𝜑 → ∀∃!𝑥(𝜓 → 𝜒))    ⇒   ⊢ (𝜑 → ∀𝑥(𝜓 → 𝜒))
Theorem	alseu2d 50758	Deduction rule: Given "all some one" applied to a top-level inference, you can extract the "exactly one" part. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (𝜑 → ∀∃!𝑥(𝜓 → 𝜒))    ⇒   ⊢ (𝜑 → ∃!𝑥𝜓)
Theorem	ralseu1d 50759	Deduction rule: Given "all some one" applied to a class, you can extract the "for all" part. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (𝜑 → ∀∃!𝑥 ∈ 𝐴(𝜓 → 𝜒))    ⇒   ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝜓 → 𝜒))
Theorem	ralseu2d 50760	Deduction rule: Given "all some one" applied to a class, you can extract the "exactly one" part. Note that the witness must satisfy the antecedent 𝜓, not merely be a member of 𝐴. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (𝜑 → ∀∃!𝑥 ∈ 𝐴(𝜓 → 𝜒))    ⇒   ⊢ (𝜑 → ∃!𝑥 ∈ 𝐴 𝜓)
Theorem	alseubii 50761	Congruence: equivalents may be substituted inside an "all some one". This is the "all some one" counterpart of alsbii 50729. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (𝜑 ↔ 𝜒)    &   ⊢ (𝜓 ↔ 𝜃)    ⇒   ⊢ (∀∃!𝑥(𝜑 → 𝜓) ↔ ∀∃!𝑥(𝜒 → 𝜃))
Theorem	ralseubii 50762	Congruence for "all some one" restricted to a class. This is the "all some one" counterpart of ralsbii 50730. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (𝜑 ↔ 𝜒)    &   ⊢ (𝜓 ↔ 𝜃)    ⇒   ⊢ (∀∃!𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ ∀∃!𝑥 ∈ 𝐴(𝜒 → 𝜃))
Theorem	nfalseu 50763*	Bound-variable hypothesis builder for "all some one". This is the "all some one" counterpart of nfals 50732. Unlike nfals 50732 it requires 𝑥 and 𝑦 to be disjoint, because the corresponding builder for ∃! is nfeuw 2618, which requires it; the version without that requirement, nfeu 2619, depends on ax-13 2401 and its use is discouraged. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ Ⅎ𝑥𝜑    &   ⊢ Ⅎ𝑥𝜓    ⇒   ⊢ Ⅎ𝑥∀∃!𝑦(𝜑 → 𝜓)
Theorem	nfralseu 50764*	Bound-variable hypothesis builder for "all some one" restricted to a class. This is the "all some one" counterpart of nfrals 50733. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ Ⅎ𝑥𝐴    &   ⊢ Ⅎ𝑥𝜑    &   ⊢ Ⅎ𝑥𝜓    ⇒   ⊢ Ⅎ𝑥∀∃!𝑦 ∈ 𝐴(𝜑 → 𝜓)
Theorem	dfalseu2 50765	An "all some one" statement is equivalent to its universal part conjoined with the claim that exactly one 𝑥 satisfies both 𝜑 and 𝜓. In other words, given ∀𝑥(𝜑 → 𝜓), requiring exactly one 𝑥 to satisfy 𝜑, which is what df-alseu 50750 requires, and requiring exactly one 𝑥 to satisfy (𝜑 ∧ 𝜓) come to the same thing. Read 𝜑 as "is a king" and 𝜓 as "is hungry": if every king is hungry, then "there is exactly one king" and "there is exactly one hungry king" say the same thing, so either of them, together with "every king is hungry", gives "the king is hungry". The universal conjunct is what makes that work, and it cannot be dropped. ∃!𝑥(𝜑 ∧ 𝜓) on its own is strictly weaker than ∀∃!𝑥(𝜑 → 𝜓), since it is satisfied when many things are 𝜑 and just one of those is 𝜓, as in a region with five kings exactly one of whom is hungry; see alseueu 50766 for the one direction that does hold without it. Uniqueness attaches to the antecedent, not to the conjunction. Russell's analysis of a definite description is built the same way: its uniqueness clause constrains the description predicate alone, while the predication is a separate conjunct. See his worked example of "the father of Charles II was executed", [Russell1905] p. 482. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (∀∃!𝑥(𝜑 → 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃!𝑥(𝜑 ∧ 𝜓)))
Theorem	alseueu 50766	"The 𝜑 is 𝜓 " implies that exactly one thing is both 𝜑 and 𝜓. This is the half of dfalseu2 50765 that drops the universal conjunct; it does not reverse, so ∃!𝑥(𝜑 ∧ 𝜓) cannot be used in place of an "all some one" statement. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ (∀∃!𝑥(𝜑 → 𝜓) → ∃!𝑥(𝜑 ∧ 𝜓))
Theorem	alseu-no-surprise 50767	Demonstrate that there is never a "surprise" when using the "all some one" quantifier, that is, it is never possible for the consequent to be both always true and always false. This follows from als-no-surprise 50735 by alseuals 50753. For a contrast, see alimp-surprise 50709. (Contributed by David A. Wheeler, 21-Jul-2026.)
⊢ ¬ (∀∃!𝑥(𝜑 → 𝜓) ∧ ∀∃!𝑥(𝜑 → ¬ 𝜓))
21.53.14  Miscellaneous Miscellaneous proofs.
Theorem	5m4e1 50768	Prove that 5 - 4 = 1. (Contributed by David A. Wheeler, 31-Jan-2017.)
⊢ (5 − 4) = 1
Theorem	2p2ne5 50769	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 50770	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 50771	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)
21.53.15  Theorems about algebraic numbers
Theorem	aacllem 50772*	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)
21.54  Mathbox for Mingli Yuan
Theorem	wrdf1d 50773	A one-to-one word maps its domain into its alphabet. (Contributed by Mingli Yuan, 11-Aug-2026.)
⊢ (𝜑 → 𝑊 ∈ Word 𝐷)    &   ⊢ (𝜑 → Fun ◡𝑊)    ⇒   ⊢ (𝜑 → 𝑊:dom 𝑊–1-1→𝐷)
21.55  Mathbox for Jiamin Zhao
21.55.1  Cross product and scalar triple product in RR^3
Theorem	1ne3 50774	1 is not equal to 3. (Contributed by Jiamin Zhao, 1-Aug-2026.)
⊢ 1 ≠ 3
Theorem	2ne3 50775	2 is not equal to 3. (Contributed by Jiamin Zhao, 1-Aug-2026.)
⊢ 2 ≠ 3
Theorem	1elfz13 50776	Membership of 1 in the integer interval ( 1 ... 3 ). (Suggested by avekens.) (Contributed by Jiamin Zhao, 1-Aug-2026.) (Proof shortened by Jiamin Zhao, 13-Aug-2026.)
⊢ 1 ∈ (1...3)
Theorem	2elfz13 50777	Membership of 2 in the integer interval ( 1 ... 3 ). (Suggested by tirix.) (Contributed by Jiamin Zhao, 1-Aug-2026.) (Proof shortened by Jiamin Zhao, 13-Aug-2026.)
⊢ 2 ∈ (1...3)
Theorem	3elfz13 50778	Membership of 3 in the integer interval ( 1 ... 3 ). (Contributed by Jiamin Zhao, 1-Aug-2026.)
⊢ 3 ∈ (1...3)
Theorem	rr3fvcl 50779	The components of a 3-dimensional real coordinate vector are real numbers. (Contributed by Jiamin Zhao, 31-Jul-2026.)
⊢ (𝐴 ∈ (ℝ ↑m (1...3)) → ((𝐴‘1) ∈ ℝ ∧ (𝐴‘2) ∈ ℝ ∧ (𝐴‘3) ∈ ℝ))
Theorem	rr3fv1cld 50780	First component of a 3-dimensional real coordinate vector is real. (Contributed by Jiamin Zhao, 10-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → (𝐴‘1) ∈ ℝ)
Theorem	rr3fv2cld 50781	Second component of a 3-dimensional real coordinate vector is real. (Contributed by Jiamin Zhao, 10-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → (𝐴‘2) ∈ ℝ)
Theorem	rr3fv3cld 50782	Third component of a 3-dimensional real coordinate vector is real. (Contributed by Jiamin Zhao, 10-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → (𝐴‘3) ∈ ℝ)
Syntax	ccrossp 50783	Extend class notation to include the cross product operation. (Contributed by Jiamin Zhao, 31-Jul-2026.)
class ⊠
Definition	df-crossp 50784*	Define the cross product of two 3-dimensional real coordinate vectors. Vectors are represented as functions on (1...3). (Contributed by Jiamin Zhao, 31-Jul-2026.)
⊢ ⊠ = (𝑢 ∈ (ℝ ↑m (1...3)), 𝑣 ∈ (ℝ ↑m (1...3)) ↦ (𝑘 ∈ (1...3) ↦ if(𝑘 = 1, (((𝑢‘2) · (𝑣‘3)) − ((𝑢‘3) · (𝑣‘2))), if(𝑘 = 2, (((𝑢‘3) · (𝑣‘1)) − ((𝑢‘1) · (𝑣‘3))), (((𝑢‘1) · (𝑣‘2)) − ((𝑢‘2) · (𝑣‘1)))))))
Syntax	ctripp 50785	Extend class notation to include the scalar triple product of 3-dimensional real coordinate vectors. (Contributed by Jiamin Zhao, 31-Jul-2026.)
class tripp
Definition	df-tripp 50786*	Define the scalar triple product of three 3-dimensional real coordinate vectors as the dot product of the first vector 𝑥 with the cross product of the other two (𝑦 and 𝑧). Apply as (𝑦(tripp‘𝑥)𝑧). Vectors are represented as functions on (1...3). (Contributed by Jiamin Zhao, 31-Jul-2026.)
⊢ tripp = (𝑥 ∈ (ℝ ↑m (1...3)) ↦ (𝑦 ∈ (ℝ ↑m (1...3)), 𝑧 ∈ (ℝ ↑m (1...3)) ↦ (ℝfld Σg (𝑘 ∈ (1...3) ↦ ((𝑥‘𝑘) · ((𝑦⊠𝑧)‘𝑘))))))
Theorem	crosspval 50787*	Value of the cross product of two 3-dimensional real coordinate vectors as a function on (1...3). (Contributed by Jiamin Zhao, 31-Jul-2026.)
⊢ ((𝐴 ∈ (ℝ ↑m (1...3)) ∧ 𝐵 ∈ (ℝ ↑m (1...3))) → (𝐴⊠𝐵) = (𝑘 ∈ (1...3) ↦ if(𝑘 = 1, (((𝐴‘2) · (𝐵‘3)) − ((𝐴‘3) · (𝐵‘2))), if(𝑘 = 2, (((𝐴‘3) · (𝐵‘1)) − ((𝐴‘1) · (𝐵‘3))), (((𝐴‘1) · (𝐵‘2)) − ((𝐴‘2) · (𝐵‘1)))))))
Theorem	crosspcle1d 50788	Closure of the first component of the cross product's coordinate formula. (Contributed by Jiamin Zhao, 11-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐵 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → (((𝐴‘2) · (𝐵‘3)) − ((𝐴‘3) · (𝐵‘2))) ∈ ℝ)
Theorem	crosspcle2d 50789	Closure of the second component of the cross product's coordinate formula. (Contributed by Jiamin Zhao, 11-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐵 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → (((𝐴‘3) · (𝐵‘1)) − ((𝐴‘1) · (𝐵‘3))) ∈ ℝ)
Theorem	crosspcle3d 50790	Closure of the third component of the cross product's coordinate formula. (Contributed by Jiamin Zhao, 11-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐵 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → (((𝐴‘1) · (𝐵‘2)) − ((𝐴‘2) · (𝐵‘1))) ∈ ℝ)
Theorem	crosspclem 50791	Lemma for crosspcld 50792. Closure of the three-way coordinate case split used in the cross product's mapping rule. (Contributed by Jiamin Zhao, 11-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐵 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → if(𝑘 = 1, (((𝐴‘2) · (𝐵‘3)) − ((𝐴‘3) · (𝐵‘2))), if(𝑘 = 2, (((𝐴‘3) · (𝐵‘1)) − ((𝐴‘1) · (𝐵‘3))), (((𝐴‘1) · (𝐵‘2)) − ((𝐴‘2) · (𝐵‘1))))) ∈ ℝ)
Theorem	crosspcld 50792	Closure of the cross product: the cross product of two 3-dimensional real coordinate vectors is again such a vector. (Contributed by Jiamin Zhao, 12-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐵 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → (𝐴⊠𝐵) ∈ (ℝ ↑m (1...3)))
Theorem	crosspv1d 50793	Value of the first component of the cross product. (Contributed by Jiamin Zhao, 12-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐵 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → ((𝐴⊠𝐵)‘1) = (((𝐴‘2) · (𝐵‘3)) − ((𝐴‘3) · (𝐵‘2))))
Theorem	crosspv2d 50794	Value of the second component of the cross product. (Contributed by Jiamin Zhao, 12-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐵 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → ((𝐴⊠𝐵)‘2) = (((𝐴‘3) · (𝐵‘1)) − ((𝐴‘1) · (𝐵‘3))))
Theorem	crosspv3d 50795	Value of the third component of the cross product. (Contributed by Jiamin Zhao, 12-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐵 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → ((𝐴⊠𝐵)‘3) = (((𝐴‘1) · (𝐵‘2)) − ((𝐴‘2) · (𝐵‘1))))
Theorem	crosspdot0lem 50796*	Lemma for crosspdotd 50798. Unfold the curried scalar triple product application into an explicit group sum. (Contributed by Jiamin Zhao, 12-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐵 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐶 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → (𝐵(tripp‘𝐴)𝐶) = (ℝfld Σg (𝑘 ∈ (1...3) ↦ ((𝐴‘𝑘) · ((𝐵⊠𝐶)‘𝑘)))))
Theorem	crosspdotsumlem 50797*	Lemma for crosspdotd 50798. Expand the group sum over (1...3) into an explicit three-term sum. (Contributed by Jiamin Zhao, 12-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐵 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐶 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → (ℝfld Σg (𝑘 ∈ (1...3) ↦ ((𝐴‘𝑘) · ((𝐵⊠𝐶)‘𝑘)))) = (((𝐴‘1) · ((𝐵⊠𝐶)‘1)) + (((𝐴‘2) · ((𝐵⊠𝐶)‘2)) + ((𝐴‘3) · ((𝐵⊠𝐶)‘3)))))
Theorem	crosspdotd 50798	Value of the scalar triple product, expanded into the standard six-term Sarrus polynomial. (Contributed by Jiamin Zhao, 12-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐵 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐶 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → (𝐵(tripp‘𝐴)𝐶) = (((((𝐴‘1) · (𝐵‘2)) · (𝐶‘3)) − (((𝐴‘1) · (𝐵‘3)) · (𝐶‘2))) + (((((𝐴‘2) · (𝐵‘3)) · (𝐶‘1)) − (((𝐴‘2) · (𝐵‘1)) · (𝐶‘3))) + ((((𝐴‘3) · (𝐵‘1)) · (𝐶‘2)) − (((𝐴‘3) · (𝐵‘2)) · (𝐶‘1))))))
Theorem	crosspaltd 50799*	Antisymmetry of the cross product: swapping the two vectors negates the result. (Contributed by Jiamin Zhao, 12-Aug-2026.)
⊢ (𝜑 → 𝐴 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝐵 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → (𝐴⊠𝐵) = (𝑘 ∈ (1...3) ↦ -((𝐵⊠𝐴)‘𝑘)))
Theorem	crossp3d 50800*	The vector triple product expansion (BAC-CAB rule): the cross product of 𝑋 with (𝑌⊠𝑍) equals 𝑌 scaled by the dot product of 𝑋 and 𝑍, minus 𝑍 scaled by the dot product of 𝑋 and 𝑌. The dot products are written out as explicit three-term sums of component products, matching the pointwise style of df-crossp 50784 rather than introducing a separate dot product operator. (Contributed by Jiamin Zhao, 12-Aug-2026.)
⊢ (𝜑 → 𝑋 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝑌 ∈ (ℝ ↑m (1...3)))    &   ⊢ (𝜑 → 𝑍 ∈ (ℝ ↑m (1...3)))    ⇒   ⊢ (𝜑 → (𝑋⊠(𝑌⊠𝑍)) = (𝑘 ∈ (1...3) ↦ ((((((𝑋‘1) · (𝑍‘1)) + ((𝑋‘2) · (𝑍‘2))) + ((𝑋‘3) · (𝑍‘3))) · (𝑌‘𝑘)) − (((((𝑋‘1) · (𝑌‘1)) + ((𝑋‘2) · (𝑌‘2))) + ((𝑋‘3) · (𝑌‘3))) · (𝑍‘𝑘)))))