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Theorem List for Metamath Proof Explorer - 42301-42360   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremonetansqsecsq 42301 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))

Theoremcotsqcscsq 42302 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))

21.36.5  Identities for "if"

Utility theorems for "if".

Theoremifnmfalse 42303 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 vs. applying iffalse 4045 directly in this case. (Contributed by David A. Wheeler, 15-May-2015.)
(𝐴𝐵 → if(𝐴𝐵, 𝐶, 𝐷) = 𝐷)

21.36.6  Decimal point

Define the decimal point operator and the decimal fraction constructor. This can model traditional decimal point notation, and serve as a convenient way to write some fractional numbers. See df-dp 42308 and df-dp2 42306 for more information; ~? dfpval provides a more convenient way to obtain a value. This is intentionally similar to df-dec 11370.

TODO: Fix non-existent label dfpval.

Syntaxcdp2 42304 Constant used for decimal fraction constructor. See df-dp2 42306.
class 𝐴𝐵

Syntaxcdp 42305 Decimal point operator. See df-dp 42308.
class .

Definitiondf-dp2 42306 Define the "decimal fraction constructor", which is used to build up "decimal fractions" in base 10. This is intentionally similar to df-dec 11370. (Contributed by David A. Wheeler, 15-May-2015.) (Revised by AV, 9-Sep-2021.)
𝐴𝐵 = (𝐴 + (𝐵 / 10))

Theoremdfdp2OLD 42307 Obsolete version of df-dp2 42306 as of 9-Sep-2021. (Contributed by David A. Wheeler, 15-May-2015.) (New usage is discouraged.) (Proof modification is discouraged.)
𝐴𝐵 = (𝐴 + (𝐵 / 10))

Definitiondf-dp 42308* Define the . (decimal point) operator. For example, (1.5) = (3 / 2), and -(32.718) = -(32718 / 1000) Unary minus, if applied, should normally be applied in front of the parentheses.

Metamath intentionally does not have a built-in construct for numbers, so it can show that numbers are something you can build based on set theory. However, that means that metamath has no built-in way to handle decimal numbers as traditionally written, e.g., "2.54", and its parsing system intentionally does not include the complexities necessary to define such a parsing system. Here we create a system for modeling traditional decimal point notation; it is not syntactically identical, but it is sufficiently similar so it is a reasonable model of decimal point notation. It should also serve as a convenient way to write some fractional numbers.

The RHS is , not ; this should simplify some proofs. The LHS is 0, since that is what is used in practice. The definition intentionally does not allow negative numbers on the LHS; if it did, nonzero fractions would produce the wrong results. (It would be possible to define the decimal point to do this, but using it would be more complicated, and the expression -(𝐴.𝐵) is just as convenient.) (Contributed by David A. Wheeler, 15-May-2015.)

. = (𝑥 ∈ ℕ0, 𝑦 ∈ ℝ ↦ 𝑥𝑦)

Theoremdp2cl 42309 Define the closure for the decimal fraction constructor if both values are reals. (Contributed by David A. Wheeler, 15-May-2015.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → 𝐴𝐵 ∈ ℝ)

Theoremdpval 42310 Define the value of the decimal point operator. See df-dp 42308. (Contributed by David A. Wheeler, 15-May-2015.)
((𝐴 ∈ ℕ0𝐵 ∈ ℝ) → (𝐴.𝐵) = 𝐴𝐵)

Theoremdpcl 42311 Prove that the closure of the decimal point is as we have defined it. See df-dp 42308. (Contributed by David A. Wheeler, 15-May-2015.)
((𝐴 ∈ ℕ0𝐵 ∈ ℝ) → (𝐴.𝐵) ∈ ℝ)

Theoremdpfrac1 42312 Prove a simple equivalence involving the decimal point. See df-dp 42308 and dpcl 42311. (Contributed by David A. Wheeler, 15-May-2015.) (Revised by AV, 9-Sep-2021.)
((𝐴 ∈ ℕ0𝐵 ∈ ℝ) → (𝐴.𝐵) = (𝐴𝐵 / 10))

Theoremdpfrac1OLD 42313 Obsolete version of dpfrac1 42312 as of 9-Sep-2021. (Contributed by David A. Wheeler, 15-May-2015.) (New usage is discouraged.) (Proof modification is discouraged.)
((𝐴 ∈ ℕ0𝐵 ∈ ℝ) → (𝐴.𝐵) = (𝐴𝐵 / 10))

21.36.7  Logarithms generalized to arbitrary base using ` logb `

Most of this subsection was moved to main set.mm, section "Logarithms to an arbitrary base".

Theoremlogb2aval 42314 Define the value of the logb function, the logarithm generalized to an arbitrary base, when used in the 2-argument form logb𝐵, 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)))

21.36.8  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 metamath as compared to the (𝐵 logb 𝑋) form defined separately.

Syntaxclog- 42315 Extend class notation to include the logarithm generalized to an arbitrary base.
class log_

Definitiondf-logbALT 42316* 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))))

21.36.9  Formally define terms 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 relationships are reflexive, but there is no formal definition of what being reflexive actually *means*. Stating the relationships directly, instead of defining a broader test such as being reflexive, can reduce proof size (because the definition of 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 whatever way we'd like.

Benoit suggests that it might be better to define these as functions. There are many advantages to doing that, but then they it 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 Benoit's approach is very much worth considering. Examples would be: BinRel = (𝑥 ∈ V ↦ {𝑟𝑟 ⊆ (𝑥 × 𝑥)}), ReflBinRel = (𝑥 ∈ V ↦ {𝑟 ∈ ( BinRel 𝑥) ∣ (Diag‘𝑥) ⊆ 𝑟}), and IrreflBinRel = (𝑥 ∈ V ↦ {𝑟 ∈ ( BinRel 𝑥) ∣ (𝑟 ∩ (Diag‘𝑥)) = ∅}).

For more discussion see: https://github.com/metamath/set.mm/pull/1286

Syntaxwreflexive 42317 Extend wff definition to include "Reflexive" applied to a class, which is true iff class R is a reflexive relationship over the set A. See df-reflexive 42318. (Contributed by David A. Wheeler, 1-Dec-2019.)
wff 𝑅Reflexive𝐴

Definitiondf-reflexive 42318* Define relexive relationship; relation R is reflexive over the set A iff 𝑥𝐴𝑥𝑅𝑥. (Contributed by David A. Wheeler, 1-Dec-2019.)
(𝑅Reflexive𝐴 ↔ (𝑅 ⊆ (𝐴 × 𝐴) ∧ ∀𝑥𝐴 𝑥𝑅𝑥))

Syntaxwirreflexive 42319 Extend wff definition to include "Irreflexive" applied to a class, which is true iff class R is an irreflexive relationship over the set A. See df-irreflexive 42320. (Contributed by David A. Wheeler, 1-Dec-2019.)
wff 𝑅Irreflexive𝐴

Definitiondf-irreflexive 42320* Define irrelexive relationship; relation R is irreflexive over the set A iff 𝑥𝐴¬ 𝑥𝑅𝑥. Note that a relationship can be neither reflexive nor irreflexive. (Contributed by David A. Wheeler, 1-Dec-2019.)
(𝑅Irreflexive𝐴 ↔ (𝑅 ⊆ (𝐴 × 𝐴) ∧ ∀𝑥𝐴 ¬ 𝑥𝑅𝑥))

21.36.10  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 10107. We might want to switch to a naming convention like addcomli 10107.

𝐵 ∈ ℂ    &   𝐶 ∈ ℂ    &   𝐴 = (𝐵 + 𝐶)       𝐴 = (𝐶 + 𝐵)

Theoremmvlladdd 42322 Move LHS left addition to RHS. (Contributed by David A. Wheeler, 15-Oct-2018.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑 → (𝐴 + 𝐵) = 𝐶)       (𝜑𝐵 = (𝐶𝐴))

Theoremmvlraddi 42323 Move LHS right addition to RHS. (Contributed by David A. Wheeler, 11-Oct-2018.)
𝐴 ∈ ℂ    &   𝐵 ∈ ℂ    &   (𝐴 + 𝐵) = 𝐶       𝐴 = (𝐶𝐵)

Theoremmvrladdd 42324 Move RHS left addition to LHS. (Contributed by David A. Wheeler, 11-Oct-2018.)
(𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴 = (𝐵 + 𝐶))       (𝜑 → (𝐴𝐵) = 𝐶)

Theoremmvrladdi 42325 Move RHS left addition to LHS. (Contributed by David A. Wheeler, 11-Oct-2018.)
𝐵 ∈ ℂ    &   𝐶 ∈ ℂ    &   𝐴 = (𝐵 + 𝐶)       (𝐴𝐵) = 𝐶

(𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝐴 = ((𝐵 + 𝐶) − 𝐷))       (𝜑𝐴 = (𝐵 + (𝐶𝐷)))

𝐵 ∈ ℂ    &   𝐶 ∈ ℂ    &   𝐷 ∈ ℂ    &   𝐴 = ((𝐵 + 𝐶) − 𝐷)       𝐴 = (𝐵 + (𝐶𝐷))

Theoremjoinlmuladdmuli 42328 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

Theoremjoinlmulsubmuld 42329 Join AB-CB into (A-C) on LHS. (Contributed by David A. Wheeler, 15-Oct-2018.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑 → ((𝐴 · 𝐵) − (𝐶 · 𝐵)) = 𝐷)       (𝜑 → ((𝐴𝐶) · 𝐵) = 𝐷)

Theoremjoinlmulsubmuli 42330 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       ((𝐴𝐶) · 𝐵) = 𝐷

Theoremmvlrmuld 42331 Move LHS right multiplication to RHS. (Contributed by David A. Wheeler, 11-Oct-2018.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐵 ≠ 0)    &   (𝜑 → (𝐴 · 𝐵) = 𝐶)       (𝜑𝐴 = (𝐶 / 𝐵))

Theoremmvlrmuli 42332 Move LHS right multiplication to RHS. (Contributed by David A. Wheeler, 11-Oct-2018.)
𝐴 ∈ ℂ    &   𝐵 ∈ ℂ    &   𝐵 ≠ 0    &   (𝐴 · 𝐵) = 𝐶       𝐴 = (𝐶 / 𝐵)

21.36.11  Algebra helper examples

Examples using the algebra helpers.

Theoremi2linesi 42333 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)    &   (𝐴𝐶) ≠ 0       𝑋 = ((𝐷𝐵) / (𝐴𝐶))

Theoremi2linesd 42334 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.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑌 = ((𝐴 · 𝑋) + 𝐵))    &   (𝜑𝑌 = ((𝐶 · 𝑋) + 𝐷))    &   (𝜑 → (𝐴𝐶) ≠ 0)       (𝜑𝑋 = ((𝐷𝐵) / (𝐴𝐶)))

21.36.12  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 42335, empty-surprise 42337, and eximp-surprise 42339.

Theoremalimp-surprise 42335 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., 𝑥𝑀 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(𝜑𝜓). 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 (𝜑𝜓) is equivalent to ¬ 𝜑𝜓 (as proven in imor 427). 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 42336) or using the allsome quantifier (df-alsi 42343) .

For other "surprises" for new users of classical logic, see empty-surprise 42337 and eximp-surprise 42339. (Contributed by David A. Wheeler, 17-Oct-2018.)

¬ ∃𝑥𝜑       (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))

Theoremalimp-no-surprise 42336 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 42335. The allsome quantifier also counters this problem, see df-alsi 42343. (Contributed by David A. Wheeler, 27-Oct-2018.)
¬ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)

Theoremempty-surprise 42337 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 𝑥𝐴𝜑 is simply an abbreviation for 𝑥(𝑥𝐴𝜑) (per df-ral 2901). Thus, if 𝐴 is the empty set, this expression is always true regardless of the value of 𝜑 (see alimp-surprise 42335).

If you want the expression 𝑥𝐴𝜑 to not be vacuously true, you need to ensure that set 𝐴 is inhabited (e.g., 𝑥𝐴). (Technical note: You can also assert that 𝐴 ≠ ∅; this is an equivalent claim in classical logic as proven in n0 3890, but in intuitionistic logic the statement 𝐴 ≠ ∅ is a weaker claim than 𝑥𝐴.)

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 non-existent entities in his logic." (Groarke, "Aristotle: Logic", section 7. (Existential Assumptions), Internet Encyclopedia of Philosophy, http://www.iep.utm.edu/aris-log/). 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 2555, celaront 2556, and cesaro 2561.

For another "surprise" for new users of classical logic, see alimp-surprise 42335 and eximp-surprise 42339. (Contributed by David A. Wheeler, 20-Oct-2018.)

¬ ∃𝑥 𝑥𝐴       𝑥𝐴 𝜑

Theoremempty-surprise2 42338 "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 42337. Here we prove an extreme example: we "prove" that false is true. Of course, we actually do no such thing (see notfal 1510); 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 42344. (Contributed by David A. Wheeler, 20-Oct-2018.)

¬ ∃𝑥 𝑥𝐴       𝑥𝐴

Theoremeximp-surprise 42339 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 427, 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 42340. See also alimp-surprise 42335 and empty-surprise 42337. (Contributed by David A. Wheeler, 17-Oct-2018.)

(∃𝑥(𝜑𝜓) ↔ ∃𝑥𝜑𝜓))

Theoremeximp-surprise2 42340 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(𝜑𝜓) 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 42339, which shows what implication really expands to. See also empty-surprise 42337. (Contributed by David A. Wheeler, 18-Oct-2018.)

𝑥 ¬ 𝜑       𝑥(𝜑𝜓)

21.36.13  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 𝑥𝜑𝜓 do not imply that 𝜑 is ever true, leading to vacuous truths. See alimp-surprise 42335 and empty-surprise 42337 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(𝜑𝜓), and when restricted (applied to a class) we allow ∀!𝑥𝐴𝜑. 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 I hope that others will eventually agree that allsome is awesome.

Syntaxwalsi 42341 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(𝜑𝜓)

Syntaxwalsc 42342 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 ∀!𝑥𝐴𝜑

Definitiondf-alsi 42343 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(𝜑𝜓) ↔ (∀𝑥(𝜑𝜓) ∧ ∃𝑥𝜑))

Definitiondf-alsc 42344 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.)
(∀!𝑥𝐴𝜑 ↔ (∀𝑥𝐴 𝜑 ∧ ∃𝑥 𝑥𝐴))

Theoremalsconv 42345 There is an equivalence between the two "all some" forms. (Contributed by David A. Wheeler, 22-Oct-2018.)
(∀!𝑥(𝑥𝐴𝜑) ↔ ∀!𝑥𝐴𝜑)

Theoremalsi1d 42346 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(𝜓𝜒))       (𝜑 → ∀𝑥(𝜓𝜒))

Theoremalsi2d 42347 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(𝜓𝜒))       (𝜑 → ∃𝑥𝜓)

Theoremalsc1d 42348 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 → ∀!𝑥𝐴𝜓)       (𝜑 → ∀𝑥𝐴 𝜓)

Theoremalsc2d 42349 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 → ∀!𝑥𝐴𝜓)       (𝜑 → ∃𝑥 𝑥𝐴)

Theoremalscn0d 42350* 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 → ∀!𝑥𝐴𝜓)       (𝜑𝐴 ≠ ∅)

Theoremalsi-no-surprise 42351 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 42343; the proof itself builds on alimp-no-surprise 42336. For a contrast, see alimp-surprise 42335. (Contributed by David A. Wheeler, 27-Oct-2018.)
¬ (∀!𝑥(𝜑𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓))

21.36.14  Miscellaneous

Miscellaneous proofs.

Theorem5m4e1 42352 Prove that 5 - 4 = 1. (Contributed by David A. Wheeler, 31-Jan-2017.)
(5 − 4) = 1

Theorem2p2ne5 42353 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

Theoremresolution 42354 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.)
(((𝜑𝜓) ∨ (¬ 𝜑𝜒)) → (𝜓𝜒))

Theoremtestable 42355 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)

21.36.15  AA theorems

Theoremaacllem 42356* 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.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝑁 ∈ ℕ0)    &   ((𝜑𝑛 ∈ (1...𝑁)) → 𝑋 ∈ ℂ)    &   ((𝜑𝑘 ∈ (0...𝑁) ∧ 𝑛 ∈ (1...𝑁)) → 𝐶 ∈ ℚ)    &   ((𝜑𝑘 ∈ (0...𝑁)) → (𝐴𝑘) = Σ𝑛 ∈ (1...𝑁)(𝐶 · 𝑋))       (𝜑𝐴 ∈ 𝔸)

21.37  Mathbox for Kunhao Zheng

21.37.1  Weighted AM-GM Inequality

Theoremamgmwlem 42357 Weighted version of amgmlem 24516. (Contributed by Kunhao Zheng, 19-Jun-2021.)
𝑀 = (mulGrp‘ℂfld)    &   (𝜑𝐴 ∈ Fin)    &   (𝜑𝐴 ≠ ∅)    &   (𝜑𝐹:𝐴⟶ℝ+)    &   (𝜑𝑊:𝐴⟶ℝ+)    &   (𝜑 → (ℂfld Σg 𝑊) = 1)       (𝜑 → (𝑀 Σg (𝐹𝑓𝑐𝑊)) ≤ (ℂfld Σg (𝐹𝑓 · 𝑊)))

TheoremamgmlemALT 42358 Alternative proof of amgmlem 24516 using amgmwlem 42357. (Proof modification is discouraged.) (New usage is discouraged.) (Contributed by Kunhao Zheng, 20-Jun-2021.)
𝑀 = (mulGrp‘ℂfld)    &   (𝜑𝐴 ∈ Fin)    &   (𝜑𝐴 ≠ ∅)    &   (𝜑𝐹:𝐴⟶ℝ+)       (𝜑 → ((𝑀 Σg 𝐹)↑𝑐(1 / (#‘𝐴))) ≤ ((ℂfld Σg 𝐹) / (#‘𝐴)))

Theoremamgmw2d 42359 Weighted arithmetic-geometric mean inequality for 𝑛 = 2 (compare amgm2d 37523). (Contributed by Kunhao Zheng, 20-Jun-2021.)
(𝜑𝐴 ∈ ℝ+)    &   (𝜑𝑃 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝑄 ∈ ℝ+)    &   (𝜑 → (𝑃 + 𝑄) = 1)       (𝜑 → ((𝐴𝑐𝑃) · (𝐵𝑐𝑄)) ≤ ((𝐴 · 𝑃) + (𝐵 · 𝑄)))

Theoremyoung2d 42360 Young's inequality for 𝑛 = 2, a direct application of amgmw2d 42359. (Contributed by Kunhao Zheng, 6-Jul-2021.)
(𝜑𝐴 ∈ ℝ+)    &   (𝜑𝑃 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝑄 ∈ ℝ+)    &   (𝜑 → ((1 / 𝑃) + (1 / 𝑄)) = 1)       (𝜑 → (𝐴 · 𝐵) ≤ (((𝐴𝑐𝑃) / 𝑃) + ((𝐵𝑐𝑄) / 𝑄)))

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