P. 268 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 26701-26800   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	relogbf 26701	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 26702	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 26703	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 26704	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 26705). (Contributed by AV, 29-Dec-2022.)
⊢ ((𝑋 ∈ (ℤ≥‘2) ∧ 𝐵 ∈ (ℤ≥‘2) ∧ (𝑋 gcd 𝐵) = 1) → (𝐵 logb 𝑋) ∈ (ℝ ∖ ℚ))
Theorem	2logb9irr 26705	Example for logbgcd1irr 26704. The logarithm of nine to base two is irrational. (Contributed by AV, 29-Dec-2022.)
⊢ (2 logb 9) ∈ (ℝ ∖ ℚ)
Theorem	logbprmirr 26706	The logarithm of a prime to a different prime base is an irrational number. For example, (2 logb 3) ∈ (ℝ ∖ ℚ) (see 2logb3irr 26707). (Contributed by AV, 31-Dec-2022.)
⊢ ((𝑋 ∈ ℙ ∧ 𝐵 ∈ ℙ ∧ 𝑋 ≠ 𝐵) → (𝐵 logb 𝑋) ∈ (ℝ ∖ ℚ))
Theorem	2logb3irr 26707	Example for logbprmirr 26706. The logarithm of three to base two is irrational. (Contributed by AV, 31-Dec-2022.)
⊢ (2 logb 3) ∈ (ℝ ∖ ℚ)
Theorem	2logb9irrALT 26708	Alternate proof of 2logb9irr 26705: 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 26709	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 16219 resp. 2logb9irr 26705), satisfying the statement in 2irrexpqALT 26710. (Contributed by AV, 29-Dec-2022.)
⊢ ((√‘2)↑𝑐(2 logb 9)) = 3
Theorem	2irrexpqALT 26710*	Alternate proof of 2irrexpq 26640: 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 26640, this is a constructive proof because it is based on two explicitly named irrational numbers (√‘2) and (2 logb 9), see sqrt2irr0 16219, 2logb9irr 26705 and sqrt2cxp2logb9e3 26709. 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 26711*	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 26712*	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 26713*	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 26714*	The (signed) angle between two vectors is the argument of their quotient. Deduction form of angval 26711. (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 26715*	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 26716*	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 26717*	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 26718*	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 26719*	Lemma for ang180 26724. Show that the "revolution number" 𝑁 is an integer, using efeq1 26437 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 26720*	Lemma for ang180 26724. 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 26478, 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 26721*	Lemma for ang180 26724. Since ang180lem1 26719 shows that 𝑁 is an integer and ang180lem2 26720 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 26722*	Lemma for ang180 26724. 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 26723*	Lemma for ang180 26724: 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 26724*	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 26725	Lemma for lawcos 26726. 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 26726*	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 26724), 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 26725 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 16115). The Pythagorean theorem pythag 26727 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 26727*	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 26724), 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 26726 to prove this, along with simple trigonometry facts like coshalfpi 26378 and cosneg 16115. (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 26728	Lemma for isosctr 26731. (Contributed by Saveliy Skresanov, 30-Dec-2016.)
⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) = 1 ∧ ¬ 1 = 𝐴) → (ℑ‘(log‘(1 − 𝐴))) ≠ π)
Theorem	isosctrlem2 26729	Lemma for isosctr 26731. 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 26730*	Lemma for isosctr 26731. 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 26731*	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 26732*	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 26726 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 26733	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 26734	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 26735	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 26736	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 26737	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 26738	Equivalence used in the proof of angpieqvd 26741. (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 26739*	Equivalence used in angpieqvd 26741. (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 26740*	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 26741*	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 26742*	If 𝑀 is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 26732 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 26743*	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 26742, where P = B, and using angrtmuld 26718 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 26744	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 26743 and the Pythagorean theorem (pythag 26727) 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 26745	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 26746	If P is on the segment AB and AQ = BQ, then PA · PB = BQ 2 − PQ 2 . This follows from two uses of chordthmlem3 26744 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 26745. (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 26747*	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 26746 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 26748*	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 26749	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 26750	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 26751	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 26752	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 26753	Lemma for dcubic1 26755 and dcubic2 26754: 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 26754*	Reverse direction of dcubic 26756. 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 26755	Forward direction of dcubic 26756: 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 26756*	Solutions to the depressed cubic, a special case of cubic 26759. (The definitions of 𝑀, 𝑁, 𝐺, 𝑇 here differ from mcubic 26757 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 26757*	Solutions to a monic cubic equation, a special case of cubic 26759. (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 26758*	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 26759*	The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 4670 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 26760	Work out a quartic binomial. (You would think that by this point it would be faster to use binom 15796, 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 26761	Lemma for dquart 26763. (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 26762	Lemma for dquart 26763. (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 26763	Solve a depressed quartic equation. To eliminate 𝑆, which is the square root of a solution 𝑀 to the resolvent cubic equation, apply cubic 26759 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 26764	Closure lemmas for quart 26771. (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 26765	Lemma for quart1 26766. (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 26766	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 26767	Lemma for quart 26771. (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 26768	Closure lemmas for quart 26771. (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 26769	Closure lemmas for quart 26771. (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 26770	Closure lemmas for quart 26771. (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 26771	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 35152) 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 26772	The arcsine function.
class arcsin
Syntax	cacos 26773	The arccosine function.
class arccos
Syntax	catan 26774	The arctangent function.
class arctan
Definition	df-asin 26775	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 26776	Define the arccosine function. See also remarks for df-asin 26775. 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 26777	Define the arctangent function. See also remarks for df-asin 26775. 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 26778	The argument to the logarithm in df-asin 26775 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → ((i · 𝐴) + (√‘(1 − (𝐴↑2)))) ≠ 0)
Theorem	asinlem2 26779	The argument to the logarithm in df-asin 26775 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 26780	Lemma for asinlem3 26781. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ ((𝐴 ∈ ℂ ∧ (ℑ‘𝐴) ≤ 0) → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))
Theorem	asinlem3 26781	The argument to the logarithm in df-asin 26775 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ (𝐴 ∈ ℂ → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))
Theorem	asinf 26782	Domain and codomain of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ arcsin:ℂ⟶ℂ
Theorem	asincl 26783	Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (arcsin‘𝐴) ∈ ℂ)
Theorem	acosf 26784	Domain and codoamin of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ arccos:ℂ⟶ℂ
Theorem	acoscl 26785	Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (arccos‘𝐴) ∈ ℂ)
Theorem	atandm 26786	Since the property is a little lengthy, we abbreviate 𝐴 ∈ ℂ ∧ 𝐴 ≠ -i ∧ 𝐴 ≠ i as 𝐴 ∈ dom arctan. This is the necessary precondition for the definition of arctan to make sense. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ 𝐴 ≠ -i ∧ 𝐴 ≠ i))
Theorem	atandm2 26787	This form of atandm 26786 is a bit more useful for showing that the logarithms in df-atan 26777 are well-defined. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 − (i · 𝐴)) ≠ 0 ∧ (1 + (i · 𝐴)) ≠ 0))
Theorem	atandm3 26788	A compact form of atandm 26786. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (𝐴↑2) ≠ -1))
Theorem	atandm4 26789	A compact form of atandm 26786. (Contributed by Mario Carneiro, 3-Apr-2015.)
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 + (𝐴↑2)) ≠ 0))
Theorem	atanf 26790	Domain and codoamin of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ arctan:(ℂ ∖ {-i, i})⟶ℂ
Theorem	atancl 26791	Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ dom arctan → (arctan‘𝐴) ∈ ℂ)
Theorem	asinval 26792	Value of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (arcsin‘𝐴) = (-i · (log‘((i · 𝐴) + (√‘(1 − (𝐴↑2)))))))
Theorem	acosval 26793	Value of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (arccos‘𝐴) = ((π / 2) − (arcsin‘𝐴)))
Theorem	atanval 26794	Value of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ dom arctan → (arctan‘𝐴) = ((i / 2) · ((log‘(1 − (i · 𝐴))) − (log‘(1 + (i · 𝐴))))))
Theorem	atanre 26795	A real number is in the domain of the arctangent function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℝ → 𝐴 ∈ dom arctan)
Theorem	asinneg 26796	The arcsine function is odd. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ (𝐴 ∈ ℂ → (arcsin‘-𝐴) = -(arcsin‘𝐴))
Theorem	acosneg 26797	The negative symmetry relation of the arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (𝐴 ∈ ℂ → (arccos‘-𝐴) = (π − (arccos‘𝐴)))
Theorem	efiasin 26798	The exponential of the arcsine function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (exp‘(i · (arcsin‘𝐴))) = ((i · 𝐴) + (√‘(1 − (𝐴↑2)))))
Theorem	sinasin 26799	The arcsine function is an inverse to sin. This is the main property that justifies the notation arcsin or sin↑-1. Because sin is not an injection, the other converse identity asinsin 26802 is only true under limited circumstances. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ (𝐴 ∈ ℂ → (sin‘(arcsin‘𝐴)) = 𝐴)
Theorem	cosacos 26800	The arccosine function is an inverse to cos. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ (𝐴 ∈ ℂ → (cos‘(arccos‘𝐴)) = 𝐴)