P. 270 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 26901-27000   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	root1eq1 26901	The only powers of an 𝑁-th root of unity that equal 1 are the multiples of 𝑁. In other words, -1↑𝑐(2 / 𝑁) has order 𝑁 in the multiplicative group of nonzero complex numbers. (In fact, these and their powers are the only elements of finite order in the complex numbers.) (Contributed by Mario Carneiro, 28-Apr-2016.)
⊢ ((𝑁 ∈ ℕ ∧ 𝐾 ∈ ℤ) → (((-1↑𝑐(2 / 𝑁))↑𝐾) = 1 ↔ 𝑁 ∥ 𝐾))
Theorem	root1cj 26902	Within the 𝑁-th roots of unity, the conjugate of the 𝐾-th root is the 𝑁 − 𝐾-th root. (Contributed by Mario Carneiro, 23-Apr-2015.)
⊢ ((𝑁 ∈ ℕ ∧ 𝐾 ∈ ℤ) → (∗‘((-1↑𝑐(2 / 𝑁))↑𝐾)) = ((-1↑𝑐(2 / 𝑁))↑(𝑁 − 𝐾)))
Theorem	cxpeq 26903*	Solve an equation involving an 𝑁-th power. The expression -1↑𝑐(2 / 𝑁) = exp(2πi / 𝑁) is a way to write the primitive 𝑁-th root of unity with the smallest positive argument. (Contributed by Mario Carneiro, 23-Apr-2015.)
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ ∧ 𝐵 ∈ ℂ) → ((𝐴↑𝑁) = 𝐵 ↔ ∃𝑛 ∈ (0...(𝑁 − 1))𝐴 = ((𝐵↑𝑐(1 / 𝑁)) · ((-1↑𝑐(2 / 𝑁))↑𝑛))))
Theorem	zrtelqelz 26904	If the 𝑁-th root of an integer 𝐴 is rational, that root is must be an integer. Similar to zsqrtelqelz 16818, generalized to positive integer roots. (Contributed by Steven Nguyen, 6-Apr-2023.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ ∧ (𝐴↑𝑐(1 / 𝑁)) ∈ ℚ) → (𝐴↑𝑐(1 / 𝑁)) ∈ ℤ)
Theorem	zrtdvds 26905	A positive integer root divides its integer. (Contributed by Steven Nguyen, 6-Apr-2023.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ ∧ (𝐴↑𝑐(1 / 𝑁)) ∈ ℕ) → (𝐴↑𝑐(1 / 𝑁)) ∥ 𝐴)
Theorem	rtprmirr 26906	The root of a prime number is irrational. (Contributed by Steven Nguyen, 6-Apr-2023.)
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ (ℤ≥‘2)) → (𝑃↑𝑐(1 / 𝑁)) ∈ (ℝ ∖ ℚ))
Theorem	loglesqrt 26907	An upper bound on the logarithm. (Contributed by Mario Carneiro, 2-May-2016.) (Proof shortened by AV, 2-Aug-2021.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (log‘(𝐴 + 1)) ≤ (√‘𝐴))
Theorem	logreclem 26908	Symmetry of the natural logarithm range by negation. Lemma for logrec 26909. (Contributed by Saveliy Skresanov, 27-Dec-2016.)
⊢ ((𝐴 ∈ ran log ∧ ¬ (ℑ‘𝐴) = π) → -𝐴 ∈ ran log)
Theorem	logrec 26909	Logarithm of a reciprocal changes sign. (Contributed by Saveliy Skresanov, 28-Dec-2016.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ (ℑ‘(log‘𝐴)) ≠ π) → (log‘𝐴) = -(log‘(1 / 𝐴)))
14.3.5  Logarithms to an arbitrary base Define "log using an arbitrary base" function and then prove some of its properties. Note that logb is generalized to an arbitrary base and arbitrary parameter in ℂ, but it doesn't accept infinities as arguments, unlike log. Metamath doesn't care what letters are used to represent classes. Usually classes begin with the letter "A", but here we use "B" and "X" to more clearly distinguish between "base" and "other parameter of log". There are different ways this could be defined in Metamath. The approach used here is intentionally similar to existing 2-parameter Metamath functions (operations): (𝐵 logb 𝑋) where 𝐵 is the base and 𝑋 is the argument of the logarithm function. An alternative would be to support the notational form (( logb ‘𝐵)‘𝑋); that looks a little more like traditional notation. Such a function ( logb ‘𝐵) for a fixed base can be obtained in Metamath (without the need for a new definition) by the curry function: (curry logb ‘𝐵), see logbmpt 26934, logbf 26935 and logbfval 26936.
Syntax	clogb 26910	Extend class notation to include the logarithm generalized to an arbitrary base.
class logb
Definition	df-logb 26911*	Define the logb operator. This is the logarithm generalized to an arbitrary base. It can be used as (𝐵 logb 𝑋) for "log base B of X". In the most common traditional notation, base B is a subscript of "log". The definition is according to Wikipedia "Complex logarithm": https://en.wikipedia.org/wiki/Complex_logarithm#Logarithms_to_other_bases (10-Jun-2020). (Contributed by David A. Wheeler, 21-Jan-2017.)
⊢ logb = (𝑥 ∈ (ℂ ∖ {0, 1}), 𝑦 ∈ (ℂ ∖ {0}) ↦ ((log‘𝑦) / (log‘𝑥)))
Theorem	logbval 26912	Define the value of the logb function, the logarithm generalized to an arbitrary base, when used as infix. Most Metamath statements select variables in order of their use, but to make the order clearer we use "B" for base and "X" for the argument of the logarithm function here. (Contributed by David A. Wheeler, 21-Jan-2017.) (Revised by David A. Wheeler, 16-Jul-2017.)
⊢ ((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ 𝑋 ∈ (ℂ ∖ {0})) → (𝐵 logb 𝑋) = ((log‘𝑋) / (log‘𝐵)))
Theorem	logbcl 26913	General logarithm closure. (Contributed by David A. Wheeler, 17-Jul-2017.)
⊢ ((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ 𝑋 ∈ (ℂ ∖ {0})) → (𝐵 logb 𝑋) ∈ ℂ)
Theorem	logbid1 26914	General logarithm is 1 when base and arg match. Property 1(a) of [Cohen4] p. 361. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by David A. Wheeler, 22-Jul-2017.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) → (𝐴 logb 𝐴) = 1)
Theorem	logb1 26915	The logarithm of 1 to an arbitrary base 𝐵 is 0. Property 1(b) of [Cohen4] p. 361. See log1 26731. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by Thierry Arnoux, 27-Sep-2017.)
⊢ ((𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ∧ 𝐵 ≠ 1) → (𝐵 logb 1) = 0)
Theorem	elogb 26916	The general logarithm of a number to the base being Euler's constant is the natural logarithm of the number. Put another way, using e as the base in logb is the same as log. Definition in [Cohen4] p. 352. (Contributed by David A. Wheeler, 17-Oct-2017.) (Revised by David A. Wheeler and AV, 16-Jun-2020.)
⊢ (𝐴 ∈ (ℂ ∖ {0}) → (e logb 𝐴) = (log‘𝐴))
Theorem	logbchbase 26917	Change of base for logarithms. Property in [Cohen4] p. 367. (Contributed by AV, 11-Jun-2020.)
⊢ (((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) ∧ (𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ∧ 𝐵 ≠ 1) ∧ 𝑋 ∈ (ℂ ∖ {0})) → (𝐴 logb 𝑋) = ((𝐵 logb 𝑋) / (𝐵 logb 𝐴)))
Theorem	relogbval 26918	Value of the general logarithm with integer base. (Contributed by Thierry Arnoux, 27-Sep-2017.)
⊢ ((𝐵 ∈ (ℤ≥‘2) ∧ 𝑋 ∈ ℝ+) → (𝐵 logb 𝑋) = ((log‘𝑋) / (log‘𝐵)))
Theorem	relogbcl 26919	Closure of the general logarithm with a positive real base on positive reals. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by Thierry Arnoux, 27-Sep-2017.)
⊢ ((𝐵 ∈ ℝ+ ∧ 𝑋 ∈ ℝ+ ∧ 𝐵 ≠ 1) → (𝐵 logb 𝑋) ∈ ℝ)
Theorem	relogbzcl 26920	Closure of the general logarithm with integer base on positive reals. (Contributed by Thierry Arnoux, 27-Sep-2017.) (Proof shortened by AV, 9-Jun-2020.)
⊢ ((𝐵 ∈ (ℤ≥‘2) ∧ 𝑋 ∈ ℝ+) → (𝐵 logb 𝑋) ∈ ℝ)
Theorem	relogbreexp 26921	Power law for the general logarithm for real powers: The logarithm of a positive real number to the power of a real number is equal to the product of the exponent and the logarithm of the base of the power. Property 4 of [Cohen4] p. 361. (Contributed by AV, 9-Jun-2020.)
⊢ ((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ 𝐶 ∈ ℝ+ ∧ 𝐸 ∈ ℝ) → (𝐵 logb (𝐶↑𝑐𝐸)) = (𝐸 · (𝐵 logb 𝐶)))
Theorem	relogbzexp 26922	Power law for the general logarithm for integer powers: The logarithm of a positive real number to the power of an integer is equal to the product of the exponent and the logarithm of the base of the power. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by AV, 9-Jun-2020.)
⊢ ((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ 𝐶 ∈ ℝ+ ∧ 𝑁 ∈ ℤ) → (𝐵 logb (𝐶↑𝑁)) = (𝑁 · (𝐵 logb 𝐶)))
Theorem	relogbmul 26923	The logarithm of the product of two positive real numbers is the sum of logarithms. Property 2 of [Cohen4] p. 361. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by AV, 29-May-2020.)
⊢ ((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ (𝐴 ∈ ℝ+ ∧ 𝐶 ∈ ℝ+)) → (𝐵 logb (𝐴 · 𝐶)) = ((𝐵 logb 𝐴) + (𝐵 logb 𝐶)))
Theorem	relogbmulexp 26924	The logarithm of the product of a positive real and a positive real number to the power of a real number is the sum of the logarithm of the first real number and the scaled logarithm of the second real number. (Contributed by AV, 29-May-2020.)
⊢ ((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ (𝐴 ∈ ℝ+ ∧ 𝐶 ∈ ℝ+ ∧ 𝐸 ∈ ℝ)) → (𝐵 logb (𝐴 · (𝐶↑𝑐𝐸))) = ((𝐵 logb 𝐴) + (𝐸 · (𝐵 logb 𝐶))))
Theorem	relogbdiv 26925	The logarithm of the quotient of two positive real numbers is the difference of logarithms. Property 3 of [Cohen4] p. 361. (Contributed by AV, 29-May-2020.)
⊢ ((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ (𝐴 ∈ ℝ+ ∧ 𝐶 ∈ ℝ+)) → (𝐵 logb (𝐴 / 𝐶)) = ((𝐵 logb 𝐴) − (𝐵 logb 𝐶)))
Theorem	relogbexp 26926	Identity law for general logarithm: the logarithm of a power to the base is the exponent. Property 6 of [Cohen4] p. 361. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by AV, 9-Jun-2020.)
⊢ ((𝐵 ∈ ℝ+ ∧ 𝐵 ≠ 1 ∧ 𝑀 ∈ ℤ) → (𝐵 logb (𝐵↑𝑀)) = 𝑀)
Theorem	nnlogbexp 26927	Identity law for general logarithm with integer base. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by Thierry Arnoux, 27-Sep-2017.)
⊢ ((𝐵 ∈ (ℤ≥‘2) ∧ 𝑀 ∈ ℤ) → (𝐵 logb (𝐵↑𝑀)) = 𝑀)
Theorem	logbrec 26928	Logarithm of a reciprocal changes sign. See logrec 26909. Particular case of Property 3 of [Cohen4] p. 361. (Contributed by Thierry Arnoux, 27-Sep-2017.)
⊢ ((𝐵 ∈ (ℤ≥‘2) ∧ 𝐴 ∈ ℝ+) → (𝐵 logb (1 / 𝐴)) = -(𝐵 logb 𝐴))
Theorem	logbleb 26929	The general logarithm function is monotone/increasing. See logleb 26749. (Contributed by Stefan O'Rear, 19-Oct-2014.) (Revised by AV, 31-May-2020.)
⊢ ((𝐵 ∈ (ℤ≥‘2) ∧ 𝑋 ∈ ℝ+ ∧ 𝑌 ∈ ℝ+) → (𝑋 ≤ 𝑌 ↔ (𝐵 logb 𝑋) ≤ (𝐵 logb 𝑌)))
Theorem	logblt 26930	The general logarithm function is strictly monotone/increasing. Property 2 of [Cohen4] p. 377. See logltb 26746. (Contributed by Stefan O'Rear, 19-Oct-2014.) (Revised by Thierry Arnoux, 27-Sep-2017.)
⊢ ((𝐵 ∈ (ℤ≥‘2) ∧ 𝑋 ∈ ℝ+ ∧ 𝑌 ∈ ℝ+) → (𝑋 < 𝑌 ↔ (𝐵 logb 𝑋) < (𝐵 logb 𝑌)))
Theorem	relogbcxp 26931	Identity law for the general logarithm for real numbers. (Contributed by AV, 22-May-2020.)
⊢ ((𝐵 ∈ (ℝ+ ∖ {1}) ∧ 𝑋 ∈ ℝ) → (𝐵 logb (𝐵↑𝑐𝑋)) = 𝑋)
Theorem	cxplogb 26932	Identity law for the general logarithm. (Contributed by AV, 22-May-2020.)
⊢ ((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ 𝑋 ∈ (ℂ ∖ {0})) → (𝐵↑𝑐(𝐵 logb 𝑋)) = 𝑋)
Theorem	relogbcxpb 26933	The logarithm is the inverse of the exponentiation. Observation in [Cohen4] p. 348. (Contributed by AV, 11-Jun-2020.)
⊢ (((𝐵 ∈ ℝ+ ∧ 𝐵 ≠ 1) ∧ 𝑋 ∈ ℝ+ ∧ 𝑌 ∈ ℝ) → ((𝐵 logb 𝑋) = 𝑌 ↔ (𝐵↑𝑐𝑌) = 𝑋))
Theorem	logbmpt 26934*	The general logarithm to a fixed base regarded as mapping. (Contributed by AV, 11-Jun-2020.)
⊢ ((𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ∧ 𝐵 ≠ 1) → (curry logb ‘𝐵) = (𝑦 ∈ (ℂ ∖ {0}) ↦ ((log‘𝑦) / (log‘𝐵))))
Theorem	logbf 26935	The general logarithm to a fixed base regarded as function. (Contributed by AV, 11-Jun-2020.)
⊢ ((𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ∧ 𝐵 ≠ 1) → (curry logb ‘𝐵):(ℂ ∖ {0})⟶ℂ)
Theorem	logbfval 26936	The general logarithm of a complex number to a fixed base. (Contributed by AV, 11-Jun-2020.)
⊢ (((𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ∧ 𝐵 ≠ 1) ∧ 𝑋 ∈ (ℂ ∖ {0})) → ((curry logb ‘𝐵)‘𝑋) = (𝐵 logb 𝑋))
Theorem	relogbf 26937	The general logarithm to a real base greater than 1 regarded as function restricted to the positive integers. Property in [Cohen4] p. 349. (Contributed by AV, 12-Jun-2020.)
⊢ ((𝐵 ∈ ℝ+ ∧ 1 < 𝐵) → ((curry logb ‘𝐵) ↾ ℝ+):ℝ+⟶ℝ)
Theorem	logblog 26938	The general logarithm to the base being Euler's constant regarded as function is the natural logarithm. (Contributed by AV, 12-Jun-2020.)
⊢ (curry logb ‘e) = log
Theorem	logbgt0b 26939	The logarithm of a positive real number to a real base greater than 1 is positive iff the number is greater than 1. (Contributed by AV, 29-Dec-2022.)
⊢ ((𝐴 ∈ ℝ+ ∧ (𝐵 ∈ ℝ+ ∧ 1 < 𝐵)) → (0 < (𝐵 logb 𝐴) ↔ 1 < 𝐴))
Theorem	logbgcd1irr 26940	The logarithm of an integer greater than 1 to an integer base greater than 1 is an irrational number if the argument and the base are relatively prime. For example, (2 logb 9) ∈ (ℝ ∖ ℚ) (see 2logb9irr 26941). (Contributed by AV, 29-Dec-2022.)
⊢ ((𝑋 ∈ (ℤ≥‘2) ∧ 𝐵 ∈ (ℤ≥‘2) ∧ (𝑋 gcd 𝐵) = 1) → (𝐵 logb 𝑋) ∈ (ℝ ∖ ℚ))
Theorem	2logb9irr 26941	Example for logbgcd1irr 26940. The logarithm of nine to base two is irrational. (Contributed by AV, 29-Dec-2022.)
⊢ (2 logb 9) ∈ (ℝ ∖ ℚ)
Theorem	logbprmirr 26942	The logarithm of a prime to a different prime base is an irrational number. For example, (2 logb 3) ∈ (ℝ ∖ ℚ) (see 2logb3irr 26943). (Contributed by AV, 31-Dec-2022.)
⊢ ((𝑋 ∈ ℙ ∧ 𝐵 ∈ ℙ ∧ 𝑋 ≠ 𝐵) → (𝐵 logb 𝑋) ∈ (ℝ ∖ ℚ))
Theorem	2logb3irr 26943	Example for logbprmirr 26942. The logarithm of three to base two is irrational. (Contributed by AV, 31-Dec-2022.)
⊢ (2 logb 3) ∈ (ℝ ∖ ℚ)
Theorem	2logb9irrALT 26944	Alternate proof of 2logb9irr 26941: The logarithm of nine to base two is irrational. (Contributed by AV, 31-Dec-2022.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (2 logb 9) ∈ (ℝ ∖ ℚ)
Theorem	sqrt2cxp2logb9e3 26945	The square root of two to the power of the logarithm of nine to base two is three. (√‘2) and (2 logb 9) are irrational numbers (see sqrt2irr0 16308 resp. 2logb9irr 26941), satisfying the statement in 2irrexpqALT 26946. (Contributed by AV, 29-Dec-2022.)
⊢ ((√‘2)↑𝑐(2 logb 9)) = 3
Theorem	2irrexpqALT 26946*	Alternate proof of 2irrexpq 26877: There exist irrational numbers 𝑎 and 𝑏 such that (𝑎↑𝑏) is rational. Statement in the Metamath book, section 1.1.5, footnote 27 on page 17, and the "constructive proof" for theorem 1.2 of [Bauer], p. 483. In contrast to 2irrexpq 26877, this is a constructive proof because it is based on two explicitly named irrational numbers (√‘2) and (2 logb 9), see sqrt2irr0 16308, 2logb9irr 26941 and sqrt2cxp2logb9e3 26945. Therefore, this proof is also acceptable/usable in intuitionistic logic. (Contributed by AV, 23-Dec-2022.) (New usage is discouraged.) (Proof modification is discouraged.)
⊢ ∃𝑎 ∈ (ℝ ∖ ℚ)∃𝑏 ∈ (ℝ ∖ ℚ)(𝑎↑𝑐𝑏) ∈ ℚ
14.3.6  Theorems of Pythagoras, isosceles triangles, and intersecting chords
Theorem	angval 26947*	Define the angle function, which takes two complex numbers, treated as vectors from the origin, and returns the angle between them, in the range ( − π, π]. To convert from the geometry notation, 𝑚𝐴𝐵𝐶, the measure of the angle with legs 𝐴𝐵, 𝐶𝐵 where 𝐶 is more counterclockwise for positive angles, is represented by ((𝐶 − 𝐵)𝐹(𝐴 − 𝐵)). (Contributed by Mario Carneiro, 23-Sep-2014.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℂ ∧ 𝐵 ≠ 0)) → (𝐴𝐹𝐵) = (ℑ‘(log‘(𝐵 / 𝐴))))
Theorem	angcan 26948*	Cancel a constant multiplier in the angle function. (Contributed by Mario Carneiro, 23-Sep-2014.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℂ ∧ 𝐵 ≠ 0) ∧ (𝐶 ∈ ℂ ∧ 𝐶 ≠ 0)) → ((𝐶 · 𝐴)𝐹(𝐶 · 𝐵)) = (𝐴𝐹𝐵))
Theorem	angneg 26949*	Cancel a negative sign in the angle function. (Contributed by Mario Carneiro, 23-Sep-2014.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℂ ∧ 𝐵 ≠ 0)) → (-𝐴𝐹-𝐵) = (𝐴𝐹𝐵))
Theorem	angvald 26950*	The (signed) angle between two vectors is the argument of their quotient. Deduction form of angval 26947. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ≠ 0)    &   ⊢ (𝜑 → 𝑌 ∈ ℂ)    &   ⊢ (𝜑 → 𝑌 ≠ 0)    ⇒   ⊢ (𝜑 → (𝑋𝐹𝑌) = (ℑ‘(log‘(𝑌 / 𝑋))))
Theorem	angcld 26951*	The (signed) angle between two vectors is in (-π(,]π). Deduction form. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ≠ 0)    &   ⊢ (𝜑 → 𝑌 ∈ ℂ)    &   ⊢ (𝜑 → 𝑌 ≠ 0)    ⇒   ⊢ (𝜑 → (𝑋𝐹𝑌) ∈ (-π(,]π))
Theorem	angrteqvd 26952*	Two vectors are at a right angle iff their quotient is purely imaginary. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ≠ 0)    &   ⊢ (𝜑 → 𝑌 ∈ ℂ)    &   ⊢ (𝜑 → 𝑌 ≠ 0)    ⇒   ⊢ (𝜑 → ((𝑋𝐹𝑌) ∈ {(π / 2), -(π / 2)} ↔ (ℜ‘(𝑌 / 𝑋)) = 0))
Theorem	cosangneg2d 26953*	The cosine of the angle between 𝑋 and -𝑌 is the negative of that between 𝑋 and 𝑌. If A, B and C are collinear points, this implies that the cosines of DBA and DBC sum to zero, i.e., that DBA and DBC are supplementary. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ≠ 0)    &   ⊢ (𝜑 → 𝑌 ∈ ℂ)    &   ⊢ (𝜑 → 𝑌 ≠ 0)    ⇒   ⊢ (𝜑 → (cos‘(𝑋𝐹-𝑌)) = -(cos‘(𝑋𝐹𝑌)))
Theorem	angrtmuld 26954*	Perpendicularity of two vectors does not change under rescaling the second. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑌 ∈ ℂ)    &   ⊢ (𝜑 → 𝑍 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ≠ 0)    &   ⊢ (𝜑 → 𝑌 ≠ 0)    &   ⊢ (𝜑 → 𝑍 ≠ 0)    &   ⊢ (𝜑 → (𝑍 / 𝑌) ∈ ℝ)    ⇒   ⊢ (𝜑 → ((𝑋𝐹𝑌) ∈ {(π / 2), -(π / 2)} ↔ (𝑋𝐹𝑍) ∈ {(π / 2), -(π / 2)}))
Theorem	ang180lem1 26955*	Lemma for ang180 26960. Show that the "revolution number" 𝑁 is an integer, using efeq1 26674 to show that since the product of the three arguments 𝐴, 1 / (1 − 𝐴), (𝐴 − 1) / 𝐴 is -1, the sum of the logarithms must be an integer multiple of 2πi away from πi = log(-1). (Contributed by Mario Carneiro, 23-Sep-2014.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ 𝑇 = (((log‘(1 / (1 − 𝐴))) + (log‘((𝐴 − 1) / 𝐴))) + (log‘𝐴))    &   ⊢ 𝑁 = (((𝑇 / i) / (2 · π)) − (1 / 2))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) → (𝑁 ∈ ℤ ∧ (𝑇 / i) ∈ ℝ))
Theorem	ang180lem2 26956*	Lemma for ang180 26960. Show that the revolution number 𝑁 is strictly between -2 and 1. Both bounds are established by iterating using the bounds on the imaginary part of the logarithm, logimcl 26715, but the resulting bound gives only 𝑁 ≤ 1 for the upper bound. The case 𝑁 = 1 is not ruled out here, but it is in some sense an "edge case" that can only happen under very specific conditions; in particular we show that all the angle arguments 𝐴, 1 / (1 − 𝐴), (𝐴 − 1) / 𝐴 must lie on the negative real axis, which is a contradiction because clearly if 𝐴 is negative then the other two are positive real. (Contributed by Mario Carneiro, 23-Sep-2014.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ 𝑇 = (((log‘(1 / (1 − 𝐴))) + (log‘((𝐴 − 1) / 𝐴))) + (log‘𝐴))    &   ⊢ 𝑁 = (((𝑇 / i) / (2 · π)) − (1 / 2))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) → (-2 < 𝑁 ∧ 𝑁 < 1))
Theorem	ang180lem3 26957*	Lemma for ang180 26960. Since ang180lem1 26955 shows that 𝑁 is an integer and ang180lem2 26956 shows that 𝑁 is strictly between -2 and 1, it follows that 𝑁 ∈ {-1, 0}, and these two cases correspond to the two possible values for 𝑇. (Contributed by Mario Carneiro, 23-Sep-2014.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ 𝑇 = (((log‘(1 / (1 − 𝐴))) + (log‘((𝐴 − 1) / 𝐴))) + (log‘𝐴))    &   ⊢ 𝑁 = (((𝑇 / i) / (2 · π)) − (1 / 2))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) → 𝑇 ∈ {-(i · π), (i · π)})
Theorem	ang180lem4 26958*	Lemma for ang180 26960. Reduce the statement to one variable. (Contributed by Mario Carneiro, 23-Sep-2014.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) → ((((1 − 𝐴)𝐹1) + (𝐴𝐹(𝐴 − 1))) + (1𝐹𝐴)) ∈ {-π, π})
Theorem	ang180lem5 26959*	Lemma for ang180 26960: Reduce the statement to two variables. (Contributed by Mario Carneiro, 23-Sep-2014.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℂ ∧ 𝐵 ≠ 0) ∧ 𝐴 ≠ 𝐵) → ((((𝐴 − 𝐵)𝐹𝐴) + (𝐵𝐹(𝐵 − 𝐴))) + (𝐴𝐹𝐵)) ∈ {-π, π})
Theorem	ang180 26960*	The sum of angles 𝑚𝐴𝐵𝐶 + 𝑚𝐵𝐶𝐴 + 𝑚𝐶𝐴𝐵 in a triangle adds up to either π or -π, i.e. 180 degrees. (The sign is due to the two possible orientations of vertex arrangement and our signed notion of angle). This is Metamath 100 proof #27. (Contributed by Mario Carneiro, 23-Sep-2014.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 𝐵 ∧ 𝐵 ≠ 𝐶 ∧ 𝐴 ≠ 𝐶)) → ((((𝐶 − 𝐵)𝐹(𝐴 − 𝐵)) + ((𝐴 − 𝐶)𝐹(𝐵 − 𝐶))) + ((𝐵 − 𝐴)𝐹(𝐶 − 𝐴))) ∈ {-π, π})
Theorem	lawcoslem1 26961	Lemma for lawcos 26962. Here we prove the law for a point at the origin and two distinct points U and V, using an expanded version of the signed angle expression on the complex plane. (Contributed by David A. Wheeler, 11-Jun-2015.)
⊢ (𝜑 → 𝑈 ∈ ℂ)    &   ⊢ (𝜑 → 𝑉 ∈ ℂ)    &   ⊢ (𝜑 → 𝑈 ≠ 0)    &   ⊢ (𝜑 → 𝑉 ≠ 0)    ⇒   ⊢ (𝜑 → ((abs‘(𝑈 − 𝑉))↑2) = ((((abs‘𝑈)↑2) + ((abs‘𝑉)↑2)) − (2 · (((abs‘𝑈) · (abs‘𝑉)) · ((ℜ‘(𝑈 / 𝑉)) / (abs‘(𝑈 / 𝑉)))))))
Theorem	lawcos 26962*	Law of cosines (also known as the Al-Kashi theorem or the generalized Pythagorean theorem, or the cosine formula or cosine rule). Given three distinct points A, B, and C, prove a relationship between their segment lengths. This theorem is expressed using the complex number plane as a plane, where 𝐹 is the signed angle construct (as used in ang180 26960), 𝑋 is the distance of line segment BC, 𝑌 is the distance of line segment AC, 𝑍 is the distance of line segment AB, and 𝑂 is the signed angle m/_ BCA on the complex plane. We translate triangle ABC to move C to the origin (C-C), B to U=(B-C), and A to V=(A-C), then use lemma lawcoslem1 26961 to prove this algebraically simpler case. The Metamath convention is to use a signed angle; in this case the sign doesn't matter because we use the cosine of the angle (see cosneg 16204). The Pythagorean theorem pythag 26963 is a special case of the law of cosines. The theorem's expression and approach were suggested by Mario Carneiro. This is Metamath 100 proof #94. (Contributed by David A. Wheeler, 12-Jun-2015.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ 𝑋 = (abs‘(𝐵 − 𝐶))    &   ⊢ 𝑌 = (abs‘(𝐴 − 𝐶))    &   ⊢ 𝑍 = (abs‘(𝐴 − 𝐵))    &   ⊢ 𝑂 = ((𝐵 − 𝐶)𝐹(𝐴 − 𝐶))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶)) → (𝑍↑2) = (((𝑋↑2) + (𝑌↑2)) − (2 · ((𝑋 · 𝑌) · (cos‘𝑂)))))
Theorem	pythag 26963*	Pythagorean theorem. Given three distinct points A, B, and C that form a right triangle (with the right angle at C), prove a relationship between their segment lengths. This theorem is expressed using the complex number plane as a plane, where 𝐹 is the signed angle construct (as used in ang180 26960), 𝑋 is the distance of line segment BC, 𝑌 is the distance of line segment AC, 𝑍 is the distance of line segment AB (the hypotenuse), and 𝑂 is the signed right angle m/_ BCA. We use the law of cosines lawcos 26962 to prove this, along with simple trigonometry facts like coshalfpi 26615 and cosneg 16204. (Contributed by David A. Wheeler, 13-Jun-2015.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ 𝑋 = (abs‘(𝐵 − 𝐶))    &   ⊢ 𝑌 = (abs‘(𝐴 − 𝐶))    &   ⊢ 𝑍 = (abs‘(𝐴 − 𝐵))    &   ⊢ 𝑂 = ((𝐵 − 𝐶)𝐹(𝐴 − 𝐶))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) ∧ 𝑂 ∈ {(π / 2), -(π / 2)}) → (𝑍↑2) = ((𝑋↑2) + (𝑌↑2)))
Theorem	isosctrlem1 26964	Lemma for isosctr 26967. (Contributed by Saveliy Skresanov, 30-Dec-2016.)
⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) = 1 ∧ ¬ 1 = 𝐴) → (ℑ‘(log‘(1 − 𝐴))) ≠ π)
Theorem	isosctrlem2 26965	Lemma for isosctr 26967. Corresponds to the case where one vertex is at 0, another at 1 and the third lies on the unit circle. (Contributed by Saveliy Skresanov, 31-Dec-2016.)
⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) = 1 ∧ ¬ 1 = 𝐴) → (ℑ‘(log‘(1 − 𝐴))) = (ℑ‘(log‘(-𝐴 / (1 − 𝐴)))))
Theorem	isosctrlem3 26966*	Lemma for isosctr 26967. Corresponds to the case where one vertex is at 0. (Contributed by Saveliy Skresanov, 1-Jan-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0 ∧ 𝐴 ≠ 𝐵) ∧ (abs‘𝐴) = (abs‘𝐵)) → (-𝐴𝐹(𝐵 − 𝐴)) = ((𝐴 − 𝐵)𝐹-𝐵))
Theorem	isosctr 26967*	Isosceles triangle theorem. This is Metamath 100 proof #65. (Contributed by Saveliy Skresanov, 1-Jan-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶 ∧ 𝐴 ≠ 𝐵) ∧ (abs‘(𝐴 − 𝐶)) = (abs‘(𝐵 − 𝐶))) → ((𝐶 − 𝐴)𝐹(𝐵 − 𝐴)) = ((𝐴 − 𝐵)𝐹(𝐶 − 𝐵)))
Theorem	ssscongptld 26968*	If two triangles have equal sides, one angle in one triangle has the same cosine as the corresponding angle in the other triangle. This is a partial form of the SSS congruence theorem. This theorem is proven by using lawcos 26962 on both triangles to express one side in terms of the other two, and then equating these expressions and reducing this algebraically to get an equality of cosines of angles. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝐸 ∈ ℂ)    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐵)    &   ⊢ (𝜑 → 𝐵 ≠ 𝐶)    &   ⊢ (𝜑 → 𝐷 ≠ 𝐸)    &   ⊢ (𝜑 → 𝐸 ≠ 𝐺)    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝐵)) = (abs‘(𝐷 − 𝐸)))    &   ⊢ (𝜑 → (abs‘(𝐵 − 𝐶)) = (abs‘(𝐸 − 𝐺)))    &   ⊢ (𝜑 → (abs‘(𝐶 − 𝐴)) = (abs‘(𝐺 − 𝐷)))    ⇒   ⊢ (𝜑 → (cos‘((𝐴 − 𝐵)𝐹(𝐶 − 𝐵))) = (cos‘((𝐷 − 𝐸)𝐹(𝐺 − 𝐸))))
Theorem	affineequiv 26969	Equivalence between two ways of expressing 𝐵 as an affine combination of 𝐴 and 𝐶. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    ⇒   ⊢ (𝜑 → (𝐵 = ((𝐷 · 𝐴) + ((1 − 𝐷) · 𝐶)) ↔ (𝐶 − 𝐵) = (𝐷 · (𝐶 − 𝐴))))
Theorem	affineequiv2 26970	Equivalence between two ways of expressing 𝐵 as an affine combination of 𝐴 and 𝐶. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    ⇒   ⊢ (𝜑 → (𝐵 = ((𝐷 · 𝐴) + ((1 − 𝐷) · 𝐶)) ↔ (𝐵 − 𝐴) = ((1 − 𝐷) · (𝐶 − 𝐴))))
Theorem	affineequiv3 26971	Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶. (Contributed by AV, 22-Jan-2023.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    ⇒   ⊢ (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ (𝐴 − 𝐵) = (𝐷 · (𝐶 − 𝐵))))
Theorem	affineequiv4 26972	Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶. (Contributed by AV, 22-Jan-2023.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    ⇒   ⊢ (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ 𝐴 = ((𝐷 · (𝐶 − 𝐵)) + 𝐵)))
Theorem	affineequivne 26973	Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶 if 𝐵 and 𝐶 are not equal. (Contributed by AV, 22-Jan-2023.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ≠ 𝐶)    ⇒   ⊢ (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ 𝐷 = ((𝐴 − 𝐵) / (𝐶 − 𝐵))))
Theorem	angpieqvdlem 26974	Equivalence used in the proof of angpieqvd 26977. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐵)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐶)    ⇒   ⊢ (𝜑 → (-((𝐶 − 𝐵) / (𝐴 − 𝐵)) ∈ ℝ+ ↔ ((𝐶 − 𝐵) / (𝐶 − 𝐴)) ∈ (0(,)1)))
Theorem	angpieqvdlem2 26975*	Equivalence used in angpieqvd 26977. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐵)    &   ⊢ (𝜑 → 𝐵 ≠ 𝐶)    ⇒   ⊢ (𝜑 → (-((𝐶 − 𝐵) / (𝐴 − 𝐵)) ∈ ℝ+ ↔ ((𝐴 − 𝐵)𝐹(𝐶 − 𝐵)) = π))
Theorem	angpined 26976*	If the angle at ABC is π, then 𝐴 is not equal to 𝐶. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐵)    &   ⊢ (𝜑 → 𝐵 ≠ 𝐶)    ⇒   ⊢ (𝜑 → (((𝐴 − 𝐵)𝐹(𝐶 − 𝐵)) = π → 𝐴 ≠ 𝐶))
Theorem	angpieqvd 26977*	The angle ABC is π iff 𝐵 is a nontrivial convex combination of 𝐴 and 𝐶, i.e., iff 𝐵 is in the interior of the segment AC. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐵)    &   ⊢ (𝜑 → 𝐵 ≠ 𝐶)    ⇒   ⊢ (𝜑 → (((𝐴 − 𝐵)𝐹(𝐶 − 𝐵)) = π ↔ ∃𝑤 ∈ (0(,)1)𝐵 = ((𝑤 · 𝐴) + ((1 − 𝑤) · 𝐶))))
Theorem	chordthmlem 26978*	If 𝑀 is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 26968 to observe that, since AMQ and BMQ have equal sides, the angles QMB and QMA must be equal. Since they are supplementary, both must be right angles. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑀 = ((𝐴 + 𝐵) / 2))    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄)))    &   ⊢ (𝜑 → 𝐴 ≠ 𝐵)    &   ⊢ (𝜑 → 𝑄 ≠ 𝑀)    ⇒   ⊢ (𝜑 → ((𝑄 − 𝑀)𝐹(𝐵 − 𝑀)) ∈ {(π / 2), -(π / 2)})
Theorem	chordthmlem2 26979*	If M is the midpoint of AB, AQ = BQ, and P is on the line AB, then QMP is a right angle. This is proven by reduction to the special case chordthmlem 26978, where P = B, and using angrtmuld 26954 to observe that QMP is right iff QMB is. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℝ)    &   ⊢ (𝜑 → 𝑀 = ((𝐴 + 𝐵) / 2))    &   ⊢ (𝜑 → 𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄)))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑀)    &   ⊢ (𝜑 → 𝑄 ≠ 𝑀)    ⇒   ⊢ (𝜑 → ((𝑄 − 𝑀)𝐹(𝑃 − 𝑀)) ∈ {(π / 2), -(π / 2)})
Theorem	chordthmlem3 26980	If M is the midpoint of AB, AQ = BQ, and P is on the line AB, then PQ 2 = QM 2 + PM 2 . This follows from chordthmlem2 26979 and the Pythagorean theorem (pythag 26963) in the case where P and Q are unequal to M. If either P or Q equals M, the result is trivial. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℝ)    &   ⊢ (𝜑 → 𝑀 = ((𝐴 + 𝐵) / 2))    &   ⊢ (𝜑 → 𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄)))    ⇒   ⊢ (𝜑 → ((abs‘(𝑃 − 𝑄))↑2) = (((abs‘(𝑄 − 𝑀))↑2) + ((abs‘(𝑃 − 𝑀))↑2)))
Theorem	chordthmlem4 26981	If P is on the segment AB and M is the midpoint of AB, then PA · PB = BM 2 − PM 2 . If all lengths are reexpressed as fractions of AB, this reduces to the identity 𝑋 · (1 − 𝑋) = (1 / 2) 2 − ((1 / 2) − 𝑋) 2 . (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ (0[,]1))    &   ⊢ (𝜑 → 𝑀 = ((𝐴 + 𝐵) / 2))    &   ⊢ (𝜑 → 𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    ⇒   ⊢ (𝜑 → ((abs‘(𝑃 − 𝐴)) · (abs‘(𝑃 − 𝐵))) = (((abs‘(𝐵 − 𝑀))↑2) − ((abs‘(𝑃 − 𝑀))↑2)))
Theorem	chordthmlem5 26982	If P is on the segment AB and AQ = BQ, then PA · PB = BQ 2 − PQ 2 . This follows from two uses of chordthmlem3 26980 to show that PQ 2 = QM 2 + PM 2 and BQ 2 = QM 2 + BM 2 , so BQ 2 − PQ 2 = (QM 2 + BM 2 ) − (QM 2 + PM 2 ) = BM 2 − PM 2 , which equals PA · PB by chordthmlem4 26981. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ (0[,]1))    &   ⊢ (𝜑 → 𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄)))    ⇒   ⊢ (𝜑 → ((abs‘(𝑃 − 𝐴)) · (abs‘(𝑃 − 𝐵))) = (((abs‘(𝐵 − 𝑄))↑2) − ((abs‘(𝑃 − 𝑄))↑2)))
Theorem	chordthm 26983*	The intersecting chords theorem. If points A, B, C, and D lie on a circle (with center Q, say), and the point P is on the interior of the segments AB and CD, then the two products of lengths PA · PB and PC · PD are equal. The Euclidean plane is identified with the complex plane, and the fact that P is on AB and on CD is expressed by the hypothesis that the angles APB and CPD are equal to π. The result is proven by using chordthmlem5 26982 twice to show that PA · PB and PC · PD both equal BQ 2 − PQ 2 . This is similar to the proof of the theorem given in Euclid's Elements, where it is Proposition III.35. This is Metamath 100 proof #55. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 𝑃)    &   ⊢ (𝜑 → 𝐵 ≠ 𝑃)    &   ⊢ (𝜑 → 𝐶 ≠ 𝑃)    &   ⊢ (𝜑 → 𝐷 ≠ 𝑃)    &   ⊢ (𝜑 → ((𝐴 − 𝑃)𝐹(𝐵 − 𝑃)) = π)    &   ⊢ (𝜑 → ((𝐶 − 𝑃)𝐹(𝐷 − 𝑃)) = π)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄)))    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐶 − 𝑄)))    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐷 − 𝑄)))    ⇒   ⊢ (𝜑 → ((abs‘(𝑃 − 𝐴)) · (abs‘(𝑃 − 𝐵))) = ((abs‘(𝑃 − 𝐶)) · (abs‘(𝑃 − 𝐷))))
Theorem	heron 26984*	Heron's formula gives the area of a triangle given only the side lengths. If points A, B, C form a triangle, then the area of the triangle, represented here as (1 / 2) · 𝑋 · 𝑌 · abs(sin𝑂), is equal to the square root of 𝑆 · (𝑆 − 𝑋) · (𝑆 − 𝑌) · (𝑆 − 𝑍), where 𝑆 = (𝑋 + 𝑌 + 𝑍) / 2 is half the perimeter of the triangle. Based on work by Jon Pennant. This is Metamath 100 proof #57. (Contributed by Mario Carneiro, 10-Mar-2019.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ 𝑋 = (abs‘(𝐵 − 𝐶))    &   ⊢ 𝑌 = (abs‘(𝐴 − 𝐶))    &   ⊢ 𝑍 = (abs‘(𝐴 − 𝐵))    &   ⊢ 𝑂 = ((𝐵 − 𝐶)𝐹(𝐴 − 𝐶))    &   ⊢ 𝑆 = (((𝑋 + 𝑌) + 𝑍) / 2)    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐶)    &   ⊢ (𝜑 → 𝐵 ≠ 𝐶)    ⇒   ⊢ (𝜑 → (((1 / 2) · (𝑋 · 𝑌)) · (abs‘(sin‘𝑂))) = (√‘((𝑆 · (𝑆 − 𝑋)) · ((𝑆 − 𝑌) · (𝑆 − 𝑍)))))
14.3.7  Solutions of quadratic, cubic, and quartic equations
Theorem	quad2 26985	The quadratic equation, without specifying the particular branch 𝐷 to the square root. (Contributed by Mario Carneiro, 23-Apr-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 0)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → (𝐷↑2) = ((𝐵↑2) − (4 · (𝐴 · 𝐶))))    ⇒   ⊢ (𝜑 → (((𝐴 · (𝑋↑2)) + ((𝐵 · 𝑋) + 𝐶)) = 0 ↔ (𝑋 = ((-𝐵 + 𝐷) / (2 · 𝐴)) ∨ 𝑋 = ((-𝐵 − 𝐷) / (2 · 𝐴)))))
Theorem	quad 26986	The quadratic equation. (Contributed by Mario Carneiro, 23-Apr-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 0)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 = ((𝐵↑2) − (4 · (𝐴 · 𝐶))))    ⇒   ⊢ (𝜑 → (((𝐴 · (𝑋↑2)) + ((𝐵 · 𝑋) + 𝐶)) = 0 ↔ (𝑋 = ((-𝐵 + (√‘𝐷)) / (2 · 𝐴)) ∨ 𝑋 = ((-𝐵 − (√‘𝐷)) / (2 · 𝐴)))))
Theorem	1cubrlem 26987	The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)
⊢ ((-1↑𝑐(2 / 3)) = ((-1 + (i · (√‘3))) / 2) ∧ ((-1↑𝑐(2 / 3))↑2) = ((-1 − (i · (√‘3))) / 2))
Theorem	1cubr 26988	The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)
⊢ 𝑅 = {1, ((-1 + (i · (√‘3))) / 2), ((-1 − (i · (√‘3))) / 2)}    ⇒   ⊢ (𝐴 ∈ 𝑅 ↔ (𝐴 ∈ ℂ ∧ (𝐴↑3) = 1))
Theorem	dcubic1lem 26989	Lemma for dcubic1 26991 and dcubic2 26990: simplify the cubic equation under the substitution 𝑋 = 𝑈 − 𝑀 / 𝑈. (Contributed by Mario Carneiro, 26-Apr-2015.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   ⊢ (𝜑 → 𝑀 = (𝑃 / 3))    &   ⊢ (𝜑 → 𝑁 = (𝑄 / 2))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑈 ∈ ℂ)    &   ⊢ (𝜑 → 𝑈 ≠ 0)    &   ⊢ (𝜑 → 𝑋 = (𝑈 − (𝑀 / 𝑈)))    ⇒   ⊢ (𝜑 → (((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0 ↔ (((𝑈↑3)↑2) + ((𝑄 · (𝑈↑3)) − (𝑀↑3))) = 0))
Theorem	dcubic2 26990*	Reverse direction of dcubic 26992. Given a solution 𝑈 to the "substitution" quadratic equation 𝑋 = 𝑈 − 𝑀 / 𝑈, show that 𝑋 is in the desired form. (Contributed by Mario Carneiro, 25-Apr-2015.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   ⊢ (𝜑 → 𝑀 = (𝑃 / 3))    &   ⊢ (𝜑 → 𝑁 = (𝑄 / 2))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑈 ∈ ℂ)    &   ⊢ (𝜑 → 𝑈 ≠ 0)    &   ⊢ (𝜑 → 𝑋 = (𝑈 − (𝑀 / 𝑈)))    &   ⊢ (𝜑 → ((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0)    ⇒   ⊢ (𝜑 → ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = ((𝑟 · 𝑇) − (𝑀 / (𝑟 · 𝑇)))))
Theorem	dcubic1 26991	Forward direction of dcubic 26992: the claimed formula produces solutions to the cubic equation. (Contributed by Mario Carneiro, 25-Apr-2015.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   ⊢ (𝜑 → 𝑀 = (𝑃 / 3))    &   ⊢ (𝜑 → 𝑁 = (𝑄 / 2))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑋 = (𝑇 − (𝑀 / 𝑇)))    ⇒   ⊢ (𝜑 → ((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0)
Theorem	dcubic 26992*	Solutions to the depressed cubic, a special case of cubic 26995. (The definitions of 𝑀, 𝑁, 𝐺, 𝑇 here differ from mcubic 26993 by scale factors of -9, 54, 54 and -27 respectively, to simplify the algebra and presentation.) (Contributed by Mario Carneiro, 26-Apr-2015.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   ⊢ (𝜑 → 𝑀 = (𝑃 / 3))    &   ⊢ (𝜑 → 𝑁 = (𝑄 / 2))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    ⇒   ⊢ (𝜑 → (((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = ((𝑟 · 𝑇) − (𝑀 / (𝑟 · 𝑇))))))
Theorem	mcubic 26993*	Solutions to a monic cubic equation, a special case of cubic 26995. (Contributed by Mario Carneiro, 24-Apr-2015.)
⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = ((𝑁 + 𝐺) / 2))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) − (4 · (𝑀↑3))))    &   ⊢ (𝜑 → 𝑀 = ((𝐵↑2) − (3 · 𝐶)))    &   ⊢ (𝜑 → 𝑁 = (((2 · (𝐵↑3)) − (9 · (𝐵 · 𝐶))) + (;27 · 𝐷)))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    ⇒   ⊢ (𝜑 → ((((𝑋↑3) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / 3))))
Theorem	cubic2 26994*	The solution to the general cubic equation, for arbitrary choices 𝐺 and 𝑇 of the square and cube roots. (Contributed by Mario Carneiro, 23-Apr-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 0)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = ((𝑁 + 𝐺) / 2))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) − (4 · (𝑀↑3))))    &   ⊢ (𝜑 → 𝑀 = ((𝐵↑2) − (3 · (𝐴 · 𝐶))))    &   ⊢ (𝜑 → 𝑁 = (((2 · (𝐵↑3)) − ((9 · 𝐴) · (𝐵 · 𝐶))) + (;27 · ((𝐴↑2) · 𝐷))))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    ⇒   ⊢ (𝜑 → ((((𝐴 · (𝑋↑3)) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / (3 · 𝐴)))))
Theorem	cubic 26995*	The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 4673 to convert the existential quantifier to a triple disjunction. This is Metamath 100 proof #37. (Contributed by Mario Carneiro, 26-Apr-2015.)
⊢ 𝑅 = {1, ((-1 + (i · (√‘3))) / 2), ((-1 − (i · (√‘3))) / 2)}    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 0)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 = (((𝑁 + (√‘𝐺)) / 2)↑𝑐(1 / 3)))    &   ⊢ (𝜑 → 𝐺 = ((𝑁↑2) − (4 · (𝑀↑3))))    &   ⊢ (𝜑 → 𝑀 = ((𝐵↑2) − (3 · (𝐴 · 𝐶))))    &   ⊢ (𝜑 → 𝑁 = (((2 · (𝐵↑3)) − ((9 · 𝐴) · (𝐵 · 𝐶))) + (;27 · ((𝐴↑2) · 𝐷))))    &   ⊢ (𝜑 → 𝑀 ≠ 0)    ⇒   ⊢ (𝜑 → ((((𝐴 · (𝑋↑3)) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟 ∈ 𝑅 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / (3 · 𝐴))))
Theorem	binom4 26996	Work out a quartic binomial. (You would think that by this point it would be faster to use binom 15886, but it turns out to be just as much work to put it into this form after clearing all the sums and calculating binomial coefficients.) (Contributed by Mario Carneiro, 6-May-2015.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐴 + 𝐵)↑4) = (((𝐴↑4) + (4 · ((𝐴↑3) · 𝐵))) + ((6 · ((𝐴↑2) · (𝐵↑2))) + ((4 · (𝐴 · (𝐵↑3))) + (𝐵↑4)))))
Theorem	dquartlem1 26997	Lemma for dquart 26999. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑆 ∈ ℂ)    &   ⊢ (𝜑 → 𝑀 = ((2 · 𝑆)↑2))    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 ∈ ℂ)    &   ⊢ (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆)))    ⇒   ⊢ (𝜑 → ((((𝑋↑2) + ((𝑀 + 𝐵) / 2)) + ((((𝑀 / 2) · 𝑋) − (𝐶 / 4)) / 𝑆)) = 0 ↔ (𝑋 = (-𝑆 + 𝐼) ∨ 𝑋 = (-𝑆 − 𝐼))))
Theorem	dquartlem2 26998	Lemma for dquart 26999. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑆 ∈ ℂ)    &   ⊢ (𝜑 → 𝑀 = ((2 · 𝑆)↑2))    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 ∈ ℂ)    &   ⊢ (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆)))    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → (((𝑀↑3) + ((2 · 𝐵) · (𝑀↑2))) + ((((𝐵↑2) − (4 · 𝐷)) · 𝑀) + -(𝐶↑2))) = 0)    ⇒   ⊢ (𝜑 → ((((𝑀 + 𝐵) / 2)↑2) − (((𝐶↑2) / 4) / 𝑀)) = 𝐷)
Theorem	dquart 26999	Solve a depressed quartic equation. To eliminate 𝑆, which is the square root of a solution 𝑀 to the resolvent cubic equation, apply cubic 26995 or one of its variants. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑆 ∈ ℂ)    &   ⊢ (𝜑 → 𝑀 = ((2 · 𝑆)↑2))    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 ∈ ℂ)    &   ⊢ (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆)))    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → (((𝑀↑3) + ((2 · 𝐵) · (𝑀↑2))) + ((((𝐵↑2) − (4 · 𝐷)) · 𝑀) + -(𝐶↑2))) = 0)    &   ⊢ (𝜑 → 𝐽 ∈ ℂ)    &   ⊢ (𝜑 → (𝐽↑2) = ((-(𝑆↑2) − (𝐵 / 2)) − ((𝐶 / 4) / 𝑆)))    ⇒   ⊢ (𝜑 → ((((𝑋↑4) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ((𝑋 = (-𝑆 + 𝐼) ∨ 𝑋 = (-𝑆 − 𝐼)) ∨ (𝑋 = (𝑆 + 𝐽) ∨ 𝑋 = (𝑆 − 𝐽)))))
Theorem	quart1cl 27000	Closure lemmas for quart 27007. (Contributed by Mario Carneiro, 7-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    ⇒   ⊢ (𝜑 → (𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ ∧ 𝑅 ∈ ℂ))