P. 259 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 25801-25900   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	dvcncxp1 25801*	Derivative of complex power with respect to first argument on the complex plane. (Contributed by Brendan Leahy, 18-Dec-2018.)
⊢ 𝐷 = (ℂ ∖ (-∞(,]0))    ⇒   ⊢ (𝐴 ∈ ℂ → (ℂ D (𝑥 ∈ 𝐷 ↦ (𝑥↑𝑐𝐴))) = (𝑥 ∈ 𝐷 ↦ (𝐴 · (𝑥↑𝑐(𝐴 − 1)))))
Theorem	dvcnsqrt 25802*	Derivative of square root function. (Contributed by Brendan Leahy, 18-Dec-2018.)
⊢ 𝐷 = (ℂ ∖ (-∞(,]0))    ⇒   ⊢ (ℂ D (𝑥 ∈ 𝐷 ↦ (√‘𝑥))) = (𝑥 ∈ 𝐷 ↦ (1 / (2 · (√‘𝑥))))
Theorem	cxpcn 25803*	Domain of continuity of the complex power function. (Contributed by Mario Carneiro, 1-May-2016.)
⊢ 𝐷 = (ℂ ∖ (-∞(,]0))    &   ⊢ 𝐽 = (TopOpen‘ℂfld)    &   ⊢ 𝐾 = (𝐽 ↾t 𝐷)    ⇒   ⊢ (𝑥 ∈ 𝐷, 𝑦 ∈ ℂ ↦ (𝑥↑𝑐𝑦)) ∈ ((𝐾 ×t 𝐽) Cn 𝐽)
Theorem	cxpcn2 25804*	Continuity of the complex power function, when the base is real. (Contributed by Mario Carneiro, 1-May-2016.)
⊢ 𝐽 = (TopOpen‘ℂfld)    &   ⊢ 𝐾 = (𝐽 ↾t ℝ+)    ⇒   ⊢ (𝑥 ∈ ℝ+, 𝑦 ∈ ℂ ↦ (𝑥↑𝑐𝑦)) ∈ ((𝐾 ×t 𝐽) Cn 𝐽)
Theorem	cxpcn3lem 25805*	Lemma for cxpcn3 25806. (Contributed by Mario Carneiro, 2-May-2016.)
⊢ 𝐷 = (◡ℜ “ ℝ+)    &   ⊢ 𝐽 = (TopOpen‘ℂfld)    &   ⊢ 𝐾 = (𝐽 ↾t (0[,)+∞))    &   ⊢ 𝐿 = (𝐽 ↾t 𝐷)    &   ⊢ 𝑈 = (if((ℜ‘𝐴) ≤ 1, (ℜ‘𝐴), 1) / 2)    &   ⊢ 𝑇 = if(𝑈 ≤ (𝐸↑𝑐(1 / 𝑈)), 𝑈, (𝐸↑𝑐(1 / 𝑈)))    ⇒   ⊢ ((𝐴 ∈ 𝐷 ∧ 𝐸 ∈ ℝ+) → ∃𝑑 ∈ ℝ+ ∀𝑎 ∈ (0[,)+∞)∀𝑏 ∈ 𝐷 (((abs‘𝑎) < 𝑑 ∧ (abs‘(𝐴 − 𝑏)) < 𝑑) → (abs‘(𝑎↑𝑐𝑏)) < 𝐸))
Theorem	cxpcn3 25806*	Extend continuity of the complex power function to a base of zero, as long as the exponent has strictly positive real part. (Contributed by Mario Carneiro, 2-May-2016.)
⊢ 𝐷 = (◡ℜ “ ℝ+)    &   ⊢ 𝐽 = (TopOpen‘ℂfld)    &   ⊢ 𝐾 = (𝐽 ↾t (0[,)+∞))    &   ⊢ 𝐿 = (𝐽 ↾t 𝐷)    ⇒   ⊢ (𝑥 ∈ (0[,)+∞), 𝑦 ∈ 𝐷 ↦ (𝑥↑𝑐𝑦)) ∈ ((𝐾 ×t 𝐿) Cn 𝐽)
Theorem	resqrtcn 25807	Continuity of the real square root function. (Contributed by Mario Carneiro, 2-May-2016.)
⊢ (√ ↾ (0[,)+∞)) ∈ ((0[,)+∞)–cn→ℝ)
Theorem	sqrtcn 25808	Continuity of the square root function. (Contributed by Mario Carneiro, 2-May-2016.)
⊢ 𝐷 = (ℂ ∖ (-∞(,]0))    ⇒   ⊢ (√ ↾ 𝐷) ∈ (𝐷–cn→ℂ)
Theorem	cxpaddlelem 25809	Lemma for cxpaddle 25810. (Contributed by Mario Carneiro, 2-Aug-2014.)
⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝐴 ≤ 1)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ+)    &   ⊢ (𝜑 → 𝐵 ≤ 1)    ⇒   ⊢ (𝜑 → 𝐴 ≤ (𝐴↑𝑐𝐵))
Theorem	cxpaddle 25810	Ordering property for complex exponentiation. (Contributed by Mario Carneiro, 8-Sep-2014.)
⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐵)    &   ⊢ (𝜑 → 𝐶 ∈ ℝ+)    &   ⊢ (𝜑 → 𝐶 ≤ 1)    ⇒   ⊢ (𝜑 → ((𝐴 + 𝐵)↑𝑐𝐶) ≤ ((𝐴↑𝑐𝐶) + (𝐵↑𝑐𝐶)))
Theorem	abscxpbnd 25811	Bound on the absolute value of a complex power. (Contributed by Mario Carneiro, 15-Sep-2014.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 0 ≤ (ℜ‘𝐵))    &   ⊢ (𝜑 → 𝑀 ∈ ℝ)    &   ⊢ (𝜑 → (abs‘𝐴) ≤ 𝑀)    ⇒   ⊢ (𝜑 → (abs‘(𝐴↑𝑐𝐵)) ≤ ((𝑀↑𝑐(ℜ‘𝐵)) · (exp‘((abs‘𝐵) · π))))
Theorem	root1id 25812	Property of an 𝑁-th root of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)
⊢ (𝑁 ∈ ℕ → ((-1↑𝑐(2 / 𝑁))↑𝑁) = 1)
Theorem	root1eq1 25813	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 25814	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 25815*	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	loglesqrt 25816	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 25817	Symmetry of the natural logarithm range by negation. Lemma for logrec 25818. (Contributed by Saveliy Skresanov, 27-Dec-2016.)
⊢ ((𝐴 ∈ ran log ∧ ¬ (ℑ‘𝐴) = π) → -𝐴 ∈ ran log)
Theorem	logrec 25818	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 25843, logbf 25844 and logbfval 25845.
Syntax	clogb 25819	Extend class notation to include the logarithm generalized to an arbitrary base.
class logb
Definition	df-logb 25820*	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 25821	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 25822	General logarithm closure. (Contributed by David A. Wheeler, 17-Jul-2017.)
⊢ ((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ 𝑋 ∈ (ℂ ∖ {0})) → (𝐵 logb 𝑋) ∈ ℂ)
Theorem	logbid1 25823	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 25824	The logarithm of 1 to an arbitrary base 𝐵 is 0. Property 1(b) of [Cohen4] p. 361. See log1 25646. (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 25825	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 25826	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 25827	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 25828	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 25829	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 25830	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 25831	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 25832	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 25833	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 25834	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 25835	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 25836	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 25837	Logarithm of a reciprocal changes sign. See logrec 25818. 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 25838	The general logarithm function is monotone/increasing. See logleb 25663. (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 25839	The general logarithm function is strictly monotone/increasing. Property 2 of [Cohen4] p. 377. See logltb 25660. (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 25840	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 25841	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 25842	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 25843*	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 25844	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 25845	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 25846	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 25847	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 25848	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 25849	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 25850). (Contributed by AV, 29-Dec-2022.)
⊢ ((𝑋 ∈ (ℤ≥‘2) ∧ 𝐵 ∈ (ℤ≥‘2) ∧ (𝑋 gcd 𝐵) = 1) → (𝐵 logb 𝑋) ∈ (ℝ ∖ ℚ))
Theorem	2logb9irr 25850	Example for logbgcd1irr 25849. The logarithm of nine to base two is irrational. (Contributed by AV, 29-Dec-2022.)
⊢ (2 logb 9) ∈ (ℝ ∖ ℚ)
Theorem	logbprmirr 25851	The logarithm of a prime to a different prime base is an irrational number. For example, (2 logb 3) ∈ (ℝ ∖ ℚ) (see 2logb3irr 25852). (Contributed by AV, 31-Dec-2022.)
⊢ ((𝑋 ∈ ℙ ∧ 𝐵 ∈ ℙ ∧ 𝑋 ≠ 𝐵) → (𝐵 logb 𝑋) ∈ (ℝ ∖ ℚ))
Theorem	2logb3irr 25852	Example for logbprmirr 25851. The logarithm of three to base two is irrational. (Contributed by AV, 31-Dec-2022.)
⊢ (2 logb 3) ∈ (ℝ ∖ ℚ)
Theorem	2logb9irrALT 25853	Alternate proof of 2logb9irr 25850: 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 25854	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 15888 resp. 2logb9irr 25850), satisfying the statement in 2irrexpqALT 25855. (Contributed by AV, 29-Dec-2022.)
⊢ ((√‘2)↑𝑐(2 logb 9)) = 3
Theorem	2irrexpqALT 25855*	Alternate proof of 2irrexpq 25790: 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 25790, this is a constructive proof because it is based on two explicitly named irrational numbers (√‘2) and (2 logb 9), see sqrt2irr0 15888, 2logb9irr 25850 and sqrt2cxp2logb9e3 25854. 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 25856*	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 25857*	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 25858*	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 25859*	The (signed) angle between two vectors is the argument of their quotient. Deduction form of angval 25856. (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 25860*	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 25861*	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 25862*	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 25863*	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 25864*	Lemma for ang180 25869. Show that the "revolution number" 𝑁 is an integer, using efeq1 25589 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 25865*	Lemma for ang180 25869. 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 25630, 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 25866*	Lemma for ang180 25869. Since ang180lem1 25864 shows that 𝑁 is an integer and ang180lem2 25865 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 25867*	Lemma for ang180 25869. 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 25868*	Lemma for ang180 25869: 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 25869*	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 25870	Lemma for lawcos 25871. 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 25871*	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 25869), 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 25870 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 15784). The Pythagorean theorem pythag 25872 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 25872*	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 25869), 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 25871 to prove this, along with simple trigonometry facts like coshalfpi 25531 and cosneg 15784. (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 25873	Lemma for isosctr 25876. (Contributed by Saveliy Skresanov, 30-Dec-2016.)
⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) = 1 ∧ ¬ 1 = 𝐴) → (ℑ‘(log‘(1 − 𝐴))) ≠ π)
Theorem	isosctrlem2 25874	Lemma for isosctr 25876. 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 25875*	Lemma for isosctr 25876. 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 25876*	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 25877*	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 25871 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 25878	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 25879	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 25880	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 25881	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 25882	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 25883	Equivalence used in the proof of angpieqvd 25886. (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 25884*	Equivalence used in angpieqvd 25886. (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 25885*	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 25886*	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 25887*	If 𝑀 is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 25877 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 25888*	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 25887, where P = B, and using angrtmuld 25863 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 25889	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 25888 and the Pythagorean theorem (pythag 25872) 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 25890	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 25891	If P is on the segment AB and AQ = BQ, then PA · PB = BQ 2 − PQ 2 . This follows from two uses of chordthmlem3 25889 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 25890. (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 25892*	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 25891 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 25893*	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 25894	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 25895	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 25896	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 25897	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 25898	Lemma for dcubic1 25900 and dcubic2 25899: 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 25899*	Reverse direction of dcubic 25901. 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 25900	Forward direction of dcubic 25901: 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)