P. 272 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 27101-27200   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	relogbval 27101	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 27102	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 27103	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 27104	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 27105	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 27106	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 27107	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 27108	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 27109	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 27110	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 27111	Logarithm of a reciprocal changes sign. See logrec 27092. 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 27112	The general logarithm function is monotone/increasing. See logleb 26932. (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 27113	The general logarithm function is strictly monotone/increasing. Property 2 of [Cohen4] p. 377. See logltb 26929. (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 27114	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 27115	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 27116	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 27117*	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 27118	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 27119	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 27120	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 27121	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 27122	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 27123	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 27124). (Contributed by AV, 29-Dec-2022.)
⊢ ((𝑋 ∈ (ℤ≥‘2) ∧ 𝐵 ∈ (ℤ≥‘2) ∧ (𝑋 gcd 𝐵) = 1) → (𝐵 logb 𝑋) ∈ (ℝ ∖ ℚ))
Theorem	2logb9irr 27124	Example for logbgcd1irr 27123. The logarithm of nine to base two is irrational. (Contributed by AV, 29-Dec-2022.)
⊢ (2 logb 9) ∈ (ℝ ∖ ℚ)
Theorem	logbprmirr 27125	The logarithm of a prime to a different prime base is an irrational number. For example, (2 logb 3) ∈ (ℝ ∖ ℚ) (see 2logb3irr 27126). (Contributed by AV, 31-Dec-2022.)
⊢ ((𝑋 ∈ ℙ ∧ 𝐵 ∈ ℙ ∧ 𝑋 ≠ 𝐵) → (𝐵 logb 𝑋) ∈ (ℝ ∖ ℚ))
Theorem	2logb3irr 27126	Example for logbprmirr 27125. The logarithm of three to base two is irrational. (Contributed by AV, 31-Dec-2022.)
⊢ (2 logb 3) ∈ (ℝ ∖ ℚ)
Theorem	2logb9irrALT 27127	Alternate proof of 2logb9irr 27124: 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 27128	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 16420 resp. 2logb9irr 27124), satisfying the statement in 2irrexpqALT 27129. (Contributed by AV, 29-Dec-2022.)
⊢ ((√‘2)↑𝑐(2 logb 9)) = 3
Theorem	2irrexpqALT 27129*	Alternate proof of 2irrexpq 27060: 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 27060, this is a constructive proof because it is based on two explicitly named irrational numbers (√‘2) and (2 logb 9), see sqrt2irr0 16420, 2logb9irr 27124 and sqrt2cxp2logb9e3 27128. 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 27130*	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 27131*	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 27132*	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 27133*	The (signed) angle between two vectors is the argument of their quotient. Deduction form of angval 27130. (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 27134*	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 27135*	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 27136*	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 27137*	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 27138*	Lemma for ang180 27143. Show that the "revolution number" 𝑁 is an integer, using efeq1 26857 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 27139*	Lemma for ang180 27143. 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 26898, 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 27140*	Lemma for ang180 27143. Since ang180lem1 27138 shows that 𝑁 is an integer and ang180lem2 27139 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 27141*	Lemma for ang180 27143. 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 27142*	Lemma for ang180 27143: 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 27143*	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 27144	Lemma for lawcos 27145. 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 27145*	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 27143), 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 27144 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 16316). The Pythagorean theorem pythag 27146 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 27146*	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 27143), 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 27145 to prove this, along with simple trigonometry facts like coshalfpi 26799 and cosneg 16316. (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 27147	Lemma for isosctr 27150. (Contributed by Saveliy Skresanov, 30-Dec-2016.)
⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) = 1 ∧ ¬ 1 = 𝐴) → (ℑ‘(log‘(1 − 𝐴))) ≠ π)
Theorem	isosctrlem2 27148	Lemma for isosctr 27150. 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 27149*	Lemma for isosctr 27150. 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 27150*	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 27151*	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 27145 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 27152	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 27153	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 27154	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 27155	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 27156	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 27157	Equivalence used in the proof of angpieqvd 27160. (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 27158*	Equivalence used in angpieqvd 27160. (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 27159*	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 27160*	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 27161*	If 𝑀 is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 27151 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 27162*	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 27161, where P = B, and using angrtmuld 27137 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 27163	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 27162 and the Pythagorean theorem (pythag 27146) 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 27164	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 27165	If P is on the segment AB and AQ = BQ, then PA · PB = BQ 2 − PQ 2 . This follows from two uses of chordthmlem3 27163 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 27164. (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 27166*	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 27165 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 27167*	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 27168	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 27169	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 27170	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 27171	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 27172	Lemma for dcubic1 27174 and dcubic2 27173: 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 27173*	Reverse direction of dcubic 27175. 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 27174	Forward direction of dcubic 27175: 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 27175*	Solutions to the depressed cubic, a special case of cubic 27178. (The definitions of 𝑀, 𝑁, 𝐺, 𝑇 here differ from mcubic 27176 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 27176*	Solutions to a monic cubic equation, a special case of cubic 27178. (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 27177*	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 27178*	The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 4666 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 27179	Work out a quartic binomial. (You would think that by this point it would be faster to use binom 16000, 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 27180	Lemma for dquart 27182. (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 27181	Lemma for dquart 27182. (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 27182	Solve a depressed quartic equation. To eliminate 𝑆, which is the square root of a solution 𝑀 to the resolvent cubic equation, apply cubic 27178 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 27183	Closure lemmas for quart 27190. (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))
Theorem	quart1lem 27184	Lemma for quart1 27185. (Contributed by Mario Carneiro, 6-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 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑌 = (𝑋 + (𝐴 / 4)))    ⇒   ⊢ (𝜑 → 𝐷 = ((((𝐴↑4) / ;;256) + (𝑃 · ((𝐴 / 4)↑2))) + ((𝑄 · (𝐴 / 4)) + 𝑅)))
Theorem	quart1 27185	Depress a quartic equation. (Contributed by Mario Carneiro, 6-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 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑌 = (𝑋 + (𝐴 / 4)))    ⇒   ⊢ (𝜑 → (((𝑋↑4) + (𝐴 · (𝑋↑3))) + ((𝐵 · (𝑋↑2)) + ((𝐶 · 𝑋) + 𝐷))) = (((𝑌↑4) + (𝑃 · (𝑌↑2))) + ((𝑄 · 𝑌) + 𝑅)))
Theorem	quartlem1 27186	Lemma for quart 27190. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑅 ∈ ℂ)    &   ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = (( -(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅))))    ⇒   ⊢ (𝜑 → (𝑈 = (((2 · 𝑃)↑2) − (3 · ((𝑃↑2) − (4 · 𝑅)))) ∧ 𝑉 = (((2 · ((2 · 𝑃)↑3)) − (9 · ((2 · 𝑃) · ((𝑃↑2) − (4 · 𝑅))))) + (;27 · -(𝑄↑2)))))
Theorem	quartlem2 27187	Closure lemmas for quart 27190. (Contributed by Mario Carneiro, 7-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐸 = -(𝐴 / 4))    &   ⊢ (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 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = (( -(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅))))    &   ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    ⇒   ⊢ (𝜑 → (𝑈 ∈ ℂ ∧ 𝑉 ∈ ℂ ∧ 𝑊 ∈ ℂ))
Theorem	quartlem3 27188	Closure lemmas for quart 27190. (Contributed by Mario Carneiro, 7-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐸 = -(𝐴 / 4))    &   ⊢ (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 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = (( -(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅))))    &   ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   ⊢ (𝜑 → 𝑆 = ((√‘𝑀) / 2))    &   ⊢ (𝜑 → 𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   ⊢ (𝜑 → 𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    ⇒   ⊢ (𝜑 → (𝑆 ∈ ℂ ∧ 𝑀 ∈ ℂ ∧ 𝑇 ∈ ℂ))
Theorem	quartlem4 27189	Closure lemmas for quart 27190. (Contributed by Mario Carneiro, 7-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐸 = -(𝐴 / 4))    &   ⊢ (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 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = (( -(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅))))    &   ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   ⊢ (𝜑 → 𝑆 = ((√‘𝑀) / 2))    &   ⊢ (𝜑 → 𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   ⊢ (𝜑 → 𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 = (√‘(( -(𝑆↑2) − (𝑃 / 2)) + ((𝑄 / 4) / 𝑆))))    &   ⊢ (𝜑 → 𝐽 = (√‘(( -(𝑆↑2) − (𝑃 / 2)) − ((𝑄 / 4) / 𝑆))))    ⇒   ⊢ (𝜑 → (𝑆 ≠ 0 ∧ 𝐼 ∈ ℂ ∧ 𝐽 ∈ ℂ))
Theorem	quart 27190	The quartic equation, writing out all roots using square and cube root functions so that only direct substitutions remain, and we can actually claim to have a "quartic equation". Naturally, this theorem is ridiculously long (see quartfull 35999) if all the substitutions are performed. This is Metamath 100 proof #46. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐸 = -(𝐴 / 4))    &   ⊢ (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 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = (( -(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅))))    &   ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   ⊢ (𝜑 → 𝑆 = ((√‘𝑀) / 2))    &   ⊢ (𝜑 → 𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   ⊢ (𝜑 → 𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 = (√‘(( -(𝑆↑2) − (𝑃 / 2)) + ((𝑄 / 4) / 𝑆))))    &   ⊢ (𝜑 → 𝐽 = (√‘(( -(𝑆↑2) − (𝑃 / 2)) − ((𝑄 / 4) / 𝑆))))    ⇒   ⊢ (𝜑 → ((((𝑋↑4) + (𝐴 · (𝑋↑3))) + ((𝐵 · (𝑋↑2)) + ((𝐶 · 𝑋) + 𝐷))) = 0 ↔ ((𝑋 = ((𝐸 − 𝑆) + 𝐼) ∨ 𝑋 = ((𝐸 − 𝑆) − 𝐼)) ∨ (𝑋 = ((𝐸 + 𝑆) + 𝐽) ∨ 𝑋 = ((𝐸 + 𝑆) − 𝐽)))))
14.3.8  Inverse trigonometric functions
Syntax	casin 27191	The arcsine function.
class arcsin
Syntax	cacos 27192	The arccosine function.
class arccos
Syntax	catan 27193	The arctangent function.
class arctan
Definition	df-asin 27194	Define the arcsine function. Because sin is not a one-to-one function, the literal inverse ◡sin is not a function. Rather than attempt to find the right domain on which to restrict sin in order to get a total function, we just define it in terms of log, which we already know is total (except at 0). There are branch points at -1 and 1 (at which the function is defined), and branch cuts along the real line not between -1 and 1, which is to say (-∞, -1) ∪ (1, +∞). (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ arcsin = (𝑥 ∈ ℂ ↦ ( -i · (log‘((i · 𝑥) + (√‘(1 − (𝑥↑2)))))))
Definition	df-acos 27195	Define the arccosine function. See also remarks for df-asin 27194. Since we define arccos in terms of arcsin, it shares the same branch points and cuts, namely (-∞, -1) ∪ (1, +∞). (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ arccos = (𝑥 ∈ ℂ ↦ ((π / 2) − (arcsin‘𝑥)))
Definition	df-atan 27196	Define the arctangent function. See also remarks for df-asin 27194. Unlike arcsin and arccos, this function is not defined everywhere, because tan(𝑧) ≠ ±i for all 𝑧 ∈ ℂ. For all other 𝑧, there is a formula for arctan(𝑧) in terms of log, and we take that as the definition. Branch points are at ±i; branch cuts are on the pure imaginary axis not between -i and i, which is to say {𝑧 ∈ ℂ ∣ (i · 𝑧) ∈ (-∞, -1) ∪ (1, +∞)}. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ arctan = (𝑥 ∈ (ℂ ∖ { -i, i}) ↦ ((i / 2) · ((log‘(1 − (i · 𝑥))) − (log‘(1 + (i · 𝑥))))))
Theorem	asinlem 27197	The argument to the logarithm in df-asin 27194 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → ((i · 𝐴) + (√‘(1 − (𝐴↑2)))) ≠ 0)
Theorem	asinlem2 27198	The argument to the logarithm in df-asin 27194 has the property that replacing 𝐴 with -𝐴 in the expression gives the reciprocal. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ (𝐴 ∈ ℂ → (((i · 𝐴) + (√‘(1 − (𝐴↑2)))) · ((i · -𝐴) + (√‘(1 − ( -𝐴↑2))))) = 1)
Theorem	asinlem3a 27199	Lemma for asinlem3 27200. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ ((𝐴 ∈ ℂ ∧ (ℑ‘𝐴) ≤ 0) → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))
Theorem	asinlem3 27200	The argument to the logarithm in df-asin 27194 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ (𝐴 ∈ ℂ → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))