P. 437 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 43601-43671   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	tanhval-named 43601	Value of the named tanh function. Here we show the simple conversion to the conventional form used in set.mm, using the definition given by df-tanh 43598. (Contributed by David A. Wheeler, 10-May-2015.)
⊢ (𝐴 ∈ (◡cosh “ (ℂ ∖ {0})) → (tanh‘𝐴) = ((tan‘(i · 𝐴)) / i))
Theorem	sinh-conventional 43602	Conventional definition of sinh. Here we show that the sinh definition we're using has the same meaning as the conventional definition used in some other sources. We choose a slightly different definition of sinh because it has fewer operations, and thus is more convenient to manipulate using set.mm. (Contributed by David A. Wheeler, 10-May-2015.)
⊢ (𝐴 ∈ ℂ → (sinh‘𝐴) = (-i · (sin‘(i · 𝐴))))
Theorem	sinhpcosh 43603	Prove that (sinh‘𝐴) + (cosh‘𝐴) = (exp‘𝐴) using the conventional hyperbolic trigonometric functions. (Contributed by David A. Wheeler, 27-May-2015.)
⊢ (𝐴 ∈ ℂ → ((sinh‘𝐴) + (cosh‘𝐴)) = (exp‘𝐴))
20.38.4  Reciprocal trigonometric functions (sec, csc, cot) Define the traditional reciprocal trigonometric functions secant (sec), cosecant (csc), and cotangent (cos), along with various identities involving them.
Syntax	csec 43604	Extend class notation to include the secant function, see df-sec 43607.
class sec
Syntax	ccsc 43605	Extend class notation to include the cosecant function, see df-csc 43608.
class csc
Syntax	ccot 43606	Extend class notation to include the cotangent function, see df-cot 43609.
class cot
Definition	df-sec 43607*	Define the secant function. We define it this way for cmpt 4967, which requires the form (𝑥 ∈ 𝐴 ↦ 𝐵). The sec function is defined in ISO 80000-2:2009(E) operation 2-13.6 and "NIST Digital Library of Mathematical Functions" section on "Trigonometric Functions" http://dlmf.nist.gov/4.14 (Contributed by David A. Wheeler, 14-Mar-2014.)
⊢ sec = (𝑥 ∈ {𝑦 ∈ ℂ ∣ (cos‘𝑦) ≠ 0} ↦ (1 / (cos‘𝑥)))
Definition	df-csc 43608*	Define the cosecant function. We define it this way for cmpt 4967, which requires the form (𝑥 ∈ 𝐴 ↦ 𝐵). The csc function is defined in ISO 80000-2:2009(E) operation 2-13.7 and "NIST Digital Library of Mathematical Functions" section on "Trigonometric Functions" http://dlmf.nist.gov/4.14 (Contributed by David A. Wheeler, 14-Mar-2014.)
⊢ csc = (𝑥 ∈ {𝑦 ∈ ℂ ∣ (sin‘𝑦) ≠ 0} ↦ (1 / (sin‘𝑥)))
Definition	df-cot 43609*	Define the cotangent function. We define it this way for cmpt 4967, which requires the form (𝑥 ∈ 𝐴 ↦ 𝐵). The cot function is defined in ISO 80000-2:2009(E) operation 2-13.5 and "NIST Digital Library of Mathematical Functions" section on "Trigonometric Functions" http://dlmf.nist.gov/4.14 (Contributed by David A. Wheeler, 14-Mar-2014.)
⊢ cot = (𝑥 ∈ {𝑦 ∈ ℂ ∣ (sin‘𝑦) ≠ 0} ↦ ((cos‘𝑥) / (sin‘𝑥)))
Theorem	secval 43610	Value of the secant function. (Contributed by David A. Wheeler, 14-Mar-2014.)
⊢ ((𝐴 ∈ ℂ ∧ (cos‘𝐴) ≠ 0) → (sec‘𝐴) = (1 / (cos‘𝐴)))
Theorem	cscval 43611	Value of the cosecant function. (Contributed by David A. Wheeler, 14-Mar-2014.)
⊢ ((𝐴 ∈ ℂ ∧ (sin‘𝐴) ≠ 0) → (csc‘𝐴) = (1 / (sin‘𝐴)))
Theorem	cotval 43612	Value of the cotangent function. (Contributed by David A. Wheeler, 14-Mar-2014.)
⊢ ((𝐴 ∈ ℂ ∧ (sin‘𝐴) ≠ 0) → (cot‘𝐴) = ((cos‘𝐴) / (sin‘𝐴)))
Theorem	seccl 43613	The closure of the secant function with a complex argument. (Contributed by David A. Wheeler, 14-Mar-2014.)
⊢ ((𝐴 ∈ ℂ ∧ (cos‘𝐴) ≠ 0) → (sec‘𝐴) ∈ ℂ)
Theorem	csccl 43614	The closure of the cosecant function with a complex argument. (Contributed by David A. Wheeler, 14-Mar-2014.)
⊢ ((𝐴 ∈ ℂ ∧ (sin‘𝐴) ≠ 0) → (csc‘𝐴) ∈ ℂ)
Theorem	cotcl 43615	The closure of the cotangent function with a complex argument. (Contributed by David A. Wheeler, 15-Mar-2014.)
⊢ ((𝐴 ∈ ℂ ∧ (sin‘𝐴) ≠ 0) → (cot‘𝐴) ∈ ℂ)
Theorem	reseccl 43616	The closure of the secant function with a real argument. (Contributed by David A. Wheeler, 15-Mar-2014.)
⊢ ((𝐴 ∈ ℝ ∧ (cos‘𝐴) ≠ 0) → (sec‘𝐴) ∈ ℝ)
Theorem	recsccl 43617	The closure of the cosecant function with a real argument. (Contributed by David A. Wheeler, 15-Mar-2014.)
⊢ ((𝐴 ∈ ℝ ∧ (sin‘𝐴) ≠ 0) → (csc‘𝐴) ∈ ℝ)
Theorem	recotcl 43618	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 43619	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 43620	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 43621	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 43622	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 43623	The value of the secant function at zero is one. (Contributed by David A. Wheeler, 16-Mar-2014.)
⊢ (sec‘0) = 1
Theorem	onetansqsecsq 43624	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 43625	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))
20.38.5  Identities for "if" Utility theorems for "if".
Theorem	ifnmfalse 43626	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 4316 directly in this case. (Contributed by David A. Wheeler, 15-May-2015.)
⊢ (𝐴 ∉ 𝐵 → if(𝐴 ∈ 𝐵, 𝐶, 𝐷) = 𝐷)
20.38.6  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 43627	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)))
20.38.7  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- 43628	Extend class notation to include the logarithm generalized to an arbitrary base.
class log_
Definition	df-logbALT 43629*	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))))
20.38.8  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 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. Benoit 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 Benoit's approach is very much worth considering. Examples would be: BinRel = (𝑥 ∈ V ↦ {𝑟 ∣ 𝑟 ⊆ (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
Syntax	wreflexive 43630	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 43631. (Contributed by David A. Wheeler, 1-Dec-2019.)
wff 𝑅Reflexive𝐴
Definition	df-reflexive 43631*	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 43632	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 43633. (Contributed by David A. Wheeler, 1-Dec-2019.)
wff 𝑅Irreflexive𝐴
Definition	df-irreflexive 43633*	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))
20.38.9  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 10570. We might want to switch to a naming convention like addcomli 10570.
Theorem	comraddi 43634	Commute RHS addition. See addcomli 10570 to commute addition on LHS. (Contributed by David A. Wheeler, 11-Oct-2018.)
⊢ 𝐵 ∈ ℂ    &   ⊢ 𝐶 ∈ ℂ    &   ⊢ 𝐴 = (𝐵 + 𝐶)    ⇒   ⊢ 𝐴 = (𝐶 + 𝐵)
Theorem	mvlladdd 43635	Move LHS left addition to RHS. (Contributed by David A. Wheeler, 15-Oct-2018.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → (𝐴 + 𝐵) = 𝐶)    ⇒   ⊢ (𝜑 → 𝐵 = (𝐶 − 𝐴))
Theorem	mvlraddi 43636	Move LHS right addition to RHS. (Contributed by David A. Wheeler, 11-Oct-2018.)
⊢ 𝐴 ∈ ℂ    &   ⊢ 𝐵 ∈ ℂ    &   ⊢ (𝐴 + 𝐵) = 𝐶    ⇒   ⊢ 𝐴 = (𝐶 − 𝐵)
Theorem	mvrladdi 43637	Move RHS left addition to LHS. (Contributed by David A. Wheeler, 11-Oct-2018.)
⊢ 𝐵 ∈ ℂ    &   ⊢ 𝐶 ∈ ℂ    &   ⊢ 𝐴 = (𝐵 + 𝐶)    ⇒   ⊢ (𝐴 − 𝐵) = 𝐶
Theorem	assraddsubi 43638	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 43639	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 43640	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 43641	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 43642	Move LHS right multiplication to RHS. (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 43643	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)
20.38.10  Algebra helper examples Examples using the algebra helpers.
Theorem	i2linesi 43644	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 43645	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)))
20.38.11  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 43646, empty-surprise 43648, and eximp-surprise 43650.
Theorem	alimp-surprise 43646	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 842). 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 43647) or using the allsome quantifier (df-alsi 43654) . For other "surprises" for new users of classical logic, see empty-surprise 43648 and eximp-surprise 43650. (Contributed by David A. Wheeler, 17-Oct-2018.)
⊢ ¬ ∃𝑥𝜑    ⇒   ⊢ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))
Theorem	alimp-no-surprise 43647	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 43646. The allsome quantifier also counters this problem, see df-alsi 43654. (Contributed by David A. Wheeler, 27-Oct-2018.)
⊢ ¬ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)
Theorem	empty-surprise 43648	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 3095). Thus, if 𝐴 is the empty set, this expression is always true regardless of the value of 𝜑 (see alimp-surprise 43646). 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 4159, 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/). 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 2701, celaront 2703, and cesaro 2712. For another "surprise" for new users of classical logic, see alimp-surprise 43646 and eximp-surprise 43650. (Contributed by David A. Wheeler, 20-Oct-2018.)
⊢ ¬ ∃𝑥 𝑥 ∈ 𝐴    ⇒   ⊢ ∀𝑥 ∈ 𝐴 𝜑
Theorem	empty-surprise2 43649	"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 43648. Here we prove an extreme example: we "prove" that false is true. Of course, we actually do no such thing (see notfal 1630); 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 43655. (Contributed by David A. Wheeler, 20-Oct-2018.)
⊢ ¬ ∃𝑥 𝑥 ∈ 𝐴    ⇒   ⊢ ∀𝑥 ∈ 𝐴 ⊥
Theorem	eximp-surprise 43650	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 842, 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 43651. See also alimp-surprise 43646 and empty-surprise 43648. (Contributed by David A. Wheeler, 17-Oct-2018.)
⊢ (∃𝑥(𝜑 → 𝜓) ↔ ∃𝑥(¬ 𝜑 ∨ 𝜓))
Theorem	eximp-surprise2 43651	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 43650, which shows what implication really expands to. See also empty-surprise 43648. (Contributed by David A. Wheeler, 18-Oct-2018.)
⊢ ∃𝑥 ¬ 𝜑    ⇒   ⊢ ∃𝑥(𝜑 → 𝜓)
20.38.12  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 43646 and empty-surprise 43648 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 I hope that others will eventually agree that allsome is awesome.
Syntax	walsi 43652	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 43653	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 43654	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 43655	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 43656	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 43657	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 43658	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 43659	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 43660	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 43661*	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 43662	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 43654; the proof itself builds on alimp-no-surprise 43647. For a contrast, see alimp-surprise 43646. (Contributed by David A. Wheeler, 27-Oct-2018.)
⊢ ¬ (∀!𝑥(𝜑 → 𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓))
20.38.13  Miscellaneous Miscellaneous proofs.
Theorem	5m4e1 43663	Prove that 5 - 4 = 1. (Contributed by David A. Wheeler, 31-Jan-2017.)
⊢ (5 − 4) = 1
Theorem	2p2ne5 43664	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 43665	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 43666	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)
20.38.14  Theorems about algebraic numbers
Theorem	aacllem 43667*	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)
20.39  Mathbox for Kunhao Zheng
20.39.1  Weighted AM-GM inequality
Theorem	amgmwlem 43668	Weighted version of amgmlem 25179. (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 43669	Alternate proof of amgmlem 25179 using amgmwlem 43668. (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 43670	Weighted arithmetic-geometric mean inequality for 𝑛 = 2 (compare amgm2d 39471). (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 43671	Young's inequality for 𝑛 = 2, a direct application of amgmw2d 43670. (Contributed by Kunhao Zheng, 6-Jul-2021.)
⊢ (𝜑 → 𝐴 ∈ ℝ+)    &   ⊢ (𝜑 → 𝑃 ∈ ℝ+)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ+)    &   ⊢ (𝜑 → 𝑄 ∈ ℝ+)    &   ⊢ (𝜑 → ((1 / 𝑃) + (1 / 𝑄)) = 1)    ⇒   ⊢ (𝜑 → (𝐴 · 𝐵) ≤ (((𝐴↑𝑐𝑃) / 𝑃) + ((𝐵↑𝑐𝑄) / 𝑄)))

< Previous  Wrap >