P. 110 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 10901-11000   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	immul2d 10901	Imaginary part of a product. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → (ℑ‘(𝐴 · 𝐵)) = (𝐴 · (ℑ‘𝐵)))
Theorem	redivapd 10902	Real part of a division. Related to remul2 10801. (Contributed by Jim Kingdon, 15-Jun-2020.)
⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 # 0)    ⇒   ⊢ (𝜑 → (ℜ‘(𝐵 / 𝐴)) = ((ℜ‘𝐵) / 𝐴))
Theorem	imdivapd 10903	Imaginary part of a division. Related to remul2 10801. (Contributed by Jim Kingdon, 15-Jun-2020.)
⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 # 0)    ⇒   ⊢ (𝜑 → (ℑ‘(𝐵 / 𝐴)) = ((ℑ‘𝐵) / 𝐴))
Theorem	crred 10904	The real part of a complex number representation. Definition 10-3.1 of [Gleason] p. 132. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ)    ⇒   ⊢ (𝜑 → (ℜ‘(𝐴 + (i · 𝐵))) = 𝐴)
Theorem	crimd 10905	The imaginary part of a complex number representation. Definition 10-3.1 of [Gleason] p. 132. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ)    ⇒   ⊢ (𝜑 → (ℑ‘(𝐴 + (i · 𝐵))) = 𝐵)
Theorem	cnreim 10906	Complex apartness in terms of real and imaginary parts. See also apreim 8492 which is similar but with different notation. (Contributed by Jim Kingdon, 16-Dec-2023.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴 # 𝐵 ↔ ((ℜ‘𝐴) # (ℜ‘𝐵) ∨ (ℑ‘𝐴) # (ℑ‘𝐵))))
4.7.3  Sequence convergence
Theorem	caucvgrelemrec 10907*	Two ways to express a reciprocal. (Contributed by Jim Kingdon, 20-Jul-2021.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐴 # 0) → (℩𝑟 ∈ ℝ (𝐴 · 𝑟) = 1) = (1 / 𝐴))
Theorem	caucvgrelemcau 10908*	Lemma for caucvgre 10909. Converting the Cauchy condition. (Contributed by Jim Kingdon, 20-Jul-2021.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ≥‘𝑛)((𝐹‘𝑛) < ((𝐹‘𝑘) + (1 / 𝑛)) ∧ (𝐹‘𝑘) < ((𝐹‘𝑛) + (1 / 𝑛))))    ⇒   ⊢ (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ ℕ (𝑛 <ℝ 𝑘 → ((𝐹‘𝑛) <ℝ ((𝐹‘𝑘) + (℩𝑟 ∈ ℝ (𝑛 · 𝑟) = 1)) ∧ (𝐹‘𝑘) <ℝ ((𝐹‘𝑛) + (℩𝑟 ∈ ℝ (𝑛 · 𝑟) = 1)))))
Theorem	caucvgre 10909*	Convergence of real sequences. A Cauchy sequence (as defined here, which has a rate of convergence built in) of real numbers converges to a real number. Specifically on rate of convergence, all terms after the nth term must be within 1 / 𝑛 of the nth term. (Contributed by Jim Kingdon, 19-Jul-2021.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ≥‘𝑛)((𝐹‘𝑛) < ((𝐹‘𝑘) + (1 / 𝑛)) ∧ (𝐹‘𝑘) < ((𝐹‘𝑛) + (1 / 𝑛))))    ⇒   ⊢ (𝜑 → ∃𝑦 ∈ ℝ ∀𝑥 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝑦 + 𝑥) ∧ 𝑦 < ((𝐹‘𝑖) + 𝑥)))
Theorem	cvg1nlemcxze 10910	Lemma for cvg1n 10914. Rearranging an expression related to the rate of convergence. (Contributed by Jim Kingdon, 6-Aug-2021.)
⊢ (𝜑 → 𝐶 ∈ ℝ+)    &   ⊢ (𝜑 → 𝑋 ∈ ℝ+)    &   ⊢ (𝜑 → 𝑍 ∈ ℕ)    &   ⊢ (𝜑 → 𝐸 ∈ ℕ)    &   ⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → ((((𝐶 · 2) / 𝑋) / 𝑍) + 𝐴) < 𝐸)    ⇒   ⊢ (𝜑 → (𝐶 / (𝐸 · 𝑍)) < (𝑋 / 2))
Theorem	cvg1nlemf 10911*	Lemma for cvg1n 10914. The modified sequence 𝐺 is a sequence. (Contributed by Jim Kingdon, 1-Aug-2021.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐶 ∈ ℝ+)    &   ⊢ (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ≥‘𝑛)((𝐹‘𝑛) < ((𝐹‘𝑘) + (𝐶 / 𝑛)) ∧ (𝐹‘𝑘) < ((𝐹‘𝑛) + (𝐶 / 𝑛))))    &   ⊢ 𝐺 = (𝑗 ∈ ℕ ↦ (𝐹‘(𝑗 · 𝑍)))    &   ⊢ (𝜑 → 𝑍 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 < 𝑍)    ⇒   ⊢ (𝜑 → 𝐺:ℕ⟶ℝ)
Theorem	cvg1nlemcau 10912*	Lemma for cvg1n 10914. By selecting spaced out terms for the modified sequence 𝐺, the terms are within 1 / 𝑛 (without the constant 𝐶). (Contributed by Jim Kingdon, 1-Aug-2021.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐶 ∈ ℝ+)    &   ⊢ (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ≥‘𝑛)((𝐹‘𝑛) < ((𝐹‘𝑘) + (𝐶 / 𝑛)) ∧ (𝐹‘𝑘) < ((𝐹‘𝑛) + (𝐶 / 𝑛))))    &   ⊢ 𝐺 = (𝑗 ∈ ℕ ↦ (𝐹‘(𝑗 · 𝑍)))    &   ⊢ (𝜑 → 𝑍 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 < 𝑍)    ⇒   ⊢ (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ≥‘𝑛)((𝐺‘𝑛) < ((𝐺‘𝑘) + (1 / 𝑛)) ∧ (𝐺‘𝑘) < ((𝐺‘𝑛) + (1 / 𝑛))))
Theorem	cvg1nlemres 10913*	Lemma for cvg1n 10914. The original sequence 𝐹 has a limit (turns out it is the same as the limit of the modified sequence 𝐺). (Contributed by Jim Kingdon, 1-Aug-2021.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐶 ∈ ℝ+)    &   ⊢ (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ≥‘𝑛)((𝐹‘𝑛) < ((𝐹‘𝑘) + (𝐶 / 𝑛)) ∧ (𝐹‘𝑘) < ((𝐹‘𝑛) + (𝐶 / 𝑛))))    &   ⊢ 𝐺 = (𝑗 ∈ ℕ ↦ (𝐹‘(𝑗 · 𝑍)))    &   ⊢ (𝜑 → 𝑍 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 < 𝑍)    ⇒   ⊢ (𝜑 → ∃𝑦 ∈ ℝ ∀𝑥 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝑦 + 𝑥) ∧ 𝑦 < ((𝐹‘𝑖) + 𝑥)))
Theorem	cvg1n 10914*	Convergence of real sequences. This is a version of caucvgre 10909 with a constant multiplier 𝐶 on the rate of convergence. That is, all terms after the nth term must be within 𝐶 / 𝑛 of the nth term. (Contributed by Jim Kingdon, 1-Aug-2021.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐶 ∈ ℝ+)    &   ⊢ (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ≥‘𝑛)((𝐹‘𝑛) < ((𝐹‘𝑘) + (𝐶 / 𝑛)) ∧ (𝐹‘𝑘) < ((𝐹‘𝑛) + (𝐶 / 𝑛))))    ⇒   ⊢ (𝜑 → ∃𝑦 ∈ ℝ ∀𝑥 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝑦 + 𝑥) ∧ 𝑦 < ((𝐹‘𝑖) + 𝑥)))
Theorem	uzin2 10915	The upper integers are closed under intersection. (Contributed by Mario Carneiro, 24-Dec-2013.)
⊢ ((𝐴 ∈ ran ℤ≥ ∧ 𝐵 ∈ ran ℤ≥) → (𝐴 ∩ 𝐵) ∈ ran ℤ≥)
Theorem	rexanuz 10916*	Combine two different upper integer properties into one. (Contributed by Mario Carneiro, 25-Dec-2013.)
⊢ (∃𝑗 ∈ ℤ ∀𝑘 ∈ (ℤ≥‘𝑗)(𝜑 ∧ 𝜓) ↔ (∃𝑗 ∈ ℤ ∀𝑘 ∈ (ℤ≥‘𝑗)𝜑 ∧ ∃𝑗 ∈ ℤ ∀𝑘 ∈ (ℤ≥‘𝑗)𝜓))
Theorem	rexfiuz 10917*	Combine finitely many different upper integer properties into one. (Contributed by Mario Carneiro, 6-Jun-2014.)
⊢ (𝐴 ∈ Fin → (∃𝑗 ∈ ℤ ∀𝑘 ∈ (ℤ≥‘𝑗)∀𝑛 ∈ 𝐴 𝜑 ↔ ∀𝑛 ∈ 𝐴 ∃𝑗 ∈ ℤ ∀𝑘 ∈ (ℤ≥‘𝑗)𝜑))
Theorem	rexuz3 10918*	Restrict the base of the upper integers set to another upper integers set. (Contributed by Mario Carneiro, 26-Dec-2013.)
⊢ 𝑍 = (ℤ≥‘𝑀)    ⇒   ⊢ (𝑀 ∈ ℤ → (∃𝑗 ∈ 𝑍 ∀𝑘 ∈ (ℤ≥‘𝑗)𝜑 ↔ ∃𝑗 ∈ ℤ ∀𝑘 ∈ (ℤ≥‘𝑗)𝜑))
Theorem	rexanuz2 10919*	Combine two different upper integer properties into one. (Contributed by Mario Carneiro, 26-Dec-2013.)
⊢ 𝑍 = (ℤ≥‘𝑀)    ⇒   ⊢ (∃𝑗 ∈ 𝑍 ∀𝑘 ∈ (ℤ≥‘𝑗)(𝜑 ∧ 𝜓) ↔ (∃𝑗 ∈ 𝑍 ∀𝑘 ∈ (ℤ≥‘𝑗)𝜑 ∧ ∃𝑗 ∈ 𝑍 ∀𝑘 ∈ (ℤ≥‘𝑗)𝜓))
Theorem	r19.29uz 10920*	A version of 19.29 1607 for upper integer quantifiers. (Contributed by Mario Carneiro, 10-Feb-2014.)
⊢ 𝑍 = (ℤ≥‘𝑀)    ⇒   ⊢ ((∀𝑘 ∈ 𝑍 𝜑 ∧ ∃𝑗 ∈ 𝑍 ∀𝑘 ∈ (ℤ≥‘𝑗)𝜓) → ∃𝑗 ∈ 𝑍 ∀𝑘 ∈ (ℤ≥‘𝑗)(𝜑 ∧ 𝜓))
Theorem	r19.2uz 10921*	A version of r19.2m 3490 for upper integer quantifiers. (Contributed by Mario Carneiro, 15-Feb-2014.)
⊢ 𝑍 = (ℤ≥‘𝑀)    ⇒   ⊢ (∃𝑗 ∈ 𝑍 ∀𝑘 ∈ (ℤ≥‘𝑗)𝜑 → ∃𝑘 ∈ 𝑍 𝜑)
Theorem	recvguniqlem 10922	Lemma for recvguniq 10923. Some of the rearrangements of the expressions. (Contributed by Jim Kingdon, 8-Aug-2021.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ)    &   ⊢ (𝜑 → 𝐾 ∈ ℕ)    &   ⊢ (𝜑 → 𝐴 < ((𝐹‘𝐾) + ((𝐴 − 𝐵) / 2)))    &   ⊢ (𝜑 → (𝐹‘𝐾) < (𝐵 + ((𝐴 − 𝐵) / 2)))    ⇒   ⊢ (𝜑 → ⊥)
Theorem	recvguniq 10923*	Limits are unique. (Contributed by Jim Kingdon, 7-Aug-2021.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐿 ∈ ℝ)    &   ⊢ (𝜑 → ∀𝑥 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑘 ∈ (ℤ≥‘𝑗)((𝐹‘𝑘) < (𝐿 + 𝑥) ∧ 𝐿 < ((𝐹‘𝑘) + 𝑥)))    &   ⊢ (𝜑 → 𝑀 ∈ ℝ)    &   ⊢ (𝜑 → ∀𝑥 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑘 ∈ (ℤ≥‘𝑗)((𝐹‘𝑘) < (𝑀 + 𝑥) ∧ 𝑀 < ((𝐹‘𝑘) + 𝑥)))    ⇒   ⊢ (𝜑 → 𝐿 = 𝑀)
4.7.4  Square root; absolute value
Syntax	csqrt 10924	Extend class notation to include square root of a complex number.
class √
Syntax	cabs 10925	Extend class notation to include a function for the absolute value (modulus) of a complex number.
class abs
Definition	df-rsqrt 10926*	Define a function whose value is the square root of a nonnegative real number. Defining the square root for complex numbers has one difficult part: choosing between the two roots. The usual way to define a principal square root for all complex numbers relies on excluded middle or something similar. But in the case of a nonnegative real number, we don't have the complications presented for general complex numbers, and we can choose the nonnegative root. (Contributed by Jim Kingdon, 23-Aug-2020.)
⊢ √ = (𝑥 ∈ ℝ ↦ (℩𝑦 ∈ ℝ ((𝑦↑2) = 𝑥 ∧ 0 ≤ 𝑦)))
Definition	df-abs 10927	Define the function for the absolute value (modulus) of a complex number. (Contributed by NM, 27-Jul-1999.)
⊢ abs = (𝑥 ∈ ℂ ↦ (√‘(𝑥 · (∗‘𝑥))))
Theorem	sqrtrval 10928*	Value of square root function. (Contributed by Jim Kingdon, 23-Aug-2020.)
⊢ (𝐴 ∈ ℝ → (√‘𝐴) = (℩𝑥 ∈ ℝ ((𝑥↑2) = 𝐴 ∧ 0 ≤ 𝑥)))
Theorem	absval 10929	The absolute value (modulus) of a complex number. Proposition 10-3.7(a) of [Gleason] p. 133. (Contributed by NM, 27-Jul-1999.) (Revised by Mario Carneiro, 7-Nov-2013.)
⊢ (𝐴 ∈ ℂ → (abs‘𝐴) = (√‘(𝐴 · (∗‘𝐴))))
Theorem	rennim 10930	A real number does not lie on the negative imaginary axis. (Contributed by Mario Carneiro, 8-Jul-2013.)
⊢ (𝐴 ∈ ℝ → (i · 𝐴) ∉ ℝ+)
Theorem	sqrt0rlem 10931	Lemma for sqrt0 10932. (Contributed by Jim Kingdon, 26-Aug-2020.)
⊢ ((𝐴 ∈ ℝ ∧ ((𝐴↑2) = 0 ∧ 0 ≤ 𝐴)) ↔ 𝐴 = 0)
Theorem	sqrt0 10932	Square root of zero. (Contributed by Mario Carneiro, 9-Jul-2013.)
⊢ (√‘0) = 0
Theorem	resqrexlem1arp 10933	Lemma for resqrex 10954. 1 + 𝐴 is a positive real (expressed in a way that will help apply seqf 10386 and similar theorems). (Contributed by Jim Kingdon, 28-Jul-2021.) (Revised by Jim Kingdon, 16-Oct-2022.)
⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → ((ℕ × {(1 + 𝐴)})‘𝑁) ∈ ℝ+)
Theorem	resqrexlemp1rp 10934*	Lemma for resqrex 10954. Applying the recursion rule yields a positive real (expressed in a way that will help apply seqf 10386 and similar theorems). (Contributed by Jim Kingdon, 28-Jul-2021.) (Revised by Jim Kingdon, 16-Oct-2022.)
⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ (𝐵 ∈ ℝ+ ∧ 𝐶 ∈ ℝ+)) → (𝐵(𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2))𝐶) ∈ ℝ+)
Theorem	resqrexlemf 10935*	Lemma for resqrex 10954. The sequence is a function. (Contributed by Mario Carneiro and Jim Kingdon, 27-Jul-2021.) (Revised by Jim Kingdon, 16-Oct-2022.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ (𝜑 → 𝐹:ℕ⟶ℝ+)
Theorem	resqrexlemf1 10936*	Lemma for resqrex 10954. Initial value. Although this sequence converges to the square root with any positive initial value, this choice makes various steps in the proof of convergence easier. (Contributed by Mario Carneiro and Jim Kingdon, 27-Jul-2021.) (Revised by Jim Kingdon, 16-Oct-2022.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ (𝜑 → (𝐹‘1) = (1 + 𝐴))
Theorem	resqrexlemfp1 10937*	Lemma for resqrex 10954. Recursion rule. This sequence is the ancient method for computing square roots, often known as the babylonian method, although known to many ancient cultures. (Contributed by Mario Carneiro and Jim Kingdon, 27-Jul-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → (𝐹‘(𝑁 + 1)) = (((𝐹‘𝑁) + (𝐴 / (𝐹‘𝑁))) / 2))
Theorem	resqrexlemover 10938*	Lemma for resqrex 10954. Each element of the sequence is an overestimate. (Contributed by Mario Carneiro and Jim Kingdon, 27-Jul-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → 𝐴 < ((𝐹‘𝑁)↑2))
Theorem	resqrexlemdec 10939*	Lemma for resqrex 10954. The sequence is decreasing. (Contributed by Mario Carneiro and Jim Kingdon, 29-Jul-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → (𝐹‘(𝑁 + 1)) < (𝐹‘𝑁))
Theorem	resqrexlemdecn 10940*	Lemma for resqrex 10954. The sequence is decreasing. (Contributed by Jim Kingdon, 31-Jul-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 < 𝑀)    ⇒   ⊢ (𝜑 → (𝐹‘𝑀) < (𝐹‘𝑁))
Theorem	resqrexlemlo 10941*	Lemma for resqrex 10954. A (variable) lower bound for each term of the sequence. (Contributed by Mario Carneiro and Jim Kingdon, 29-Jul-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → (1 / (2↑𝑁)) < (𝐹‘𝑁))
Theorem	resqrexlemcalc1 10942*	Lemma for resqrex 10954. Some of the calculations involved in showing that the sequence converges. (Contributed by Mario Carneiro and Jim Kingdon, 29-Jul-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → (((𝐹‘(𝑁 + 1))↑2) − 𝐴) = (((((𝐹‘𝑁)↑2) − 𝐴)↑2) / (4 · ((𝐹‘𝑁)↑2))))
Theorem	resqrexlemcalc2 10943*	Lemma for resqrex 10954. Some of the calculations involved in showing that the sequence converges. (Contributed by Mario Carneiro and Jim Kingdon, 29-Jul-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → (((𝐹‘(𝑁 + 1))↑2) − 𝐴) ≤ ((((𝐹‘𝑁)↑2) − 𝐴) / 4))
Theorem	resqrexlemcalc3 10944*	Lemma for resqrex 10954. Some of the calculations involved in showing that the sequence converges. (Contributed by Mario Carneiro and Jim Kingdon, 29-Jul-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → (((𝐹‘𝑁)↑2) − 𝐴) ≤ (((𝐹‘1)↑2) / (4↑(𝑁 − 1))))
Theorem	resqrexlemnmsq 10945*	Lemma for resqrex 10954. The difference between the squares of two terms of the sequence. (Contributed by Mario Carneiro and Jim Kingdon, 30-Jul-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ≤ 𝑀)    ⇒   ⊢ (𝜑 → (((𝐹‘𝑁)↑2) − ((𝐹‘𝑀)↑2)) < (((𝐹‘1)↑2) / (4↑(𝑁 − 1))))
Theorem	resqrexlemnm 10946*	Lemma for resqrex 10954. The difference between two terms of the sequence. (Contributed by Mario Carneiro and Jim Kingdon, 31-Jul-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ≤ 𝑀)    ⇒   ⊢ (𝜑 → ((𝐹‘𝑁) − (𝐹‘𝑀)) < ((((𝐹‘1)↑2) · 2) / (2↑(𝑁 − 1))))
Theorem	resqrexlemcvg 10947*	Lemma for resqrex 10954. The sequence has a limit. (Contributed by Jim Kingdon, 6-Aug-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ (𝜑 → ∃𝑟 ∈ ℝ ∀𝑥 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝑟 + 𝑥) ∧ 𝑟 < ((𝐹‘𝑖) + 𝑥)))
Theorem	resqrexlemgt0 10948*	Lemma for resqrex 10954. A limit is nonnegative. (Contributed by Jim Kingdon, 7-Aug-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝐿 ∈ ℝ)    &   ⊢ (𝜑 → ∀𝑒 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹‘𝑖) + 𝑒)))    ⇒   ⊢ (𝜑 → 0 ≤ 𝐿)
Theorem	resqrexlemoverl 10949*	Lemma for resqrex 10954. Every term in the sequence is an overestimate compared with the limit 𝐿. Although this theorem is stated in terms of a particular sequence the proof could be adapted for any decreasing convergent sequence. (Contributed by Jim Kingdon, 9-Aug-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝐿 ∈ ℝ)    &   ⊢ (𝜑 → ∀𝑒 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹‘𝑖) + 𝑒)))    &   ⊢ (𝜑 → 𝐾 ∈ ℕ)    ⇒   ⊢ (𝜑 → 𝐿 ≤ (𝐹‘𝐾))
Theorem	resqrexlemglsq 10950*	Lemma for resqrex 10954. The sequence formed by squaring each term of 𝐹 converges to (𝐿↑2). (Contributed by Mario Carneiro and Jim Kingdon, 8-Aug-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝐿 ∈ ℝ)    &   ⊢ (𝜑 → ∀𝑒 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹‘𝑖) + 𝑒)))    &   ⊢ 𝐺 = (𝑥 ∈ ℕ ↦ ((𝐹‘𝑥)↑2))    ⇒   ⊢ (𝜑 → ∀𝑒 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑘 ∈ (ℤ≥‘𝑗)((𝐺‘𝑘) < ((𝐿↑2) + 𝑒) ∧ (𝐿↑2) < ((𝐺‘𝑘) + 𝑒)))
Theorem	resqrexlemga 10951*	Lemma for resqrex 10954. The sequence formed by squaring each term of 𝐹 converges to 𝐴. (Contributed by Mario Carneiro and Jim Kingdon, 8-Aug-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝐿 ∈ ℝ)    &   ⊢ (𝜑 → ∀𝑒 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹‘𝑖) + 𝑒)))    &   ⊢ 𝐺 = (𝑥 ∈ ℕ ↦ ((𝐹‘𝑥)↑2))    ⇒   ⊢ (𝜑 → ∀𝑒 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑘 ∈ (ℤ≥‘𝑗)((𝐺‘𝑘) < (𝐴 + 𝑒) ∧ 𝐴 < ((𝐺‘𝑘) + 𝑒)))
Theorem	resqrexlemsqa 10952*	Lemma for resqrex 10954. The square of a limit is 𝐴. (Contributed by Jim Kingdon, 7-Aug-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝐿 ∈ ℝ)    &   ⊢ (𝜑 → ∀𝑒 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹‘𝑖) + 𝑒)))    ⇒   ⊢ (𝜑 → (𝐿↑2) = 𝐴)
Theorem	resqrexlemex 10953*	Lemma for resqrex 10954. Existence of square root given a sequence which converges to the square root. (Contributed by Mario Carneiro and Jim Kingdon, 27-Jul-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ (𝜑 → ∃𝑥 ∈ ℝ (0 ≤ 𝑥 ∧ (𝑥↑2) = 𝐴))
Theorem	resqrex 10954*	Existence of a square root for positive reals. (Contributed by Mario Carneiro, 9-Jul-2013.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ∃𝑥 ∈ ℝ (0 ≤ 𝑥 ∧ (𝑥↑2) = 𝐴))
Theorem	rsqrmo 10955*	Uniqueness for the square root function. (Contributed by Jim Kingdon, 10-Aug-2021.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ∃*𝑥 ∈ ℝ ((𝑥↑2) = 𝐴 ∧ 0 ≤ 𝑥))
Theorem	rersqreu 10956*	Existence and uniqueness for the real square root function. (Contributed by Jim Kingdon, 10-Aug-2021.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ∃!𝑥 ∈ ℝ ((𝑥↑2) = 𝐴 ∧ 0 ≤ 𝑥))
Theorem	resqrtcl 10957	Closure of the square root function. (Contributed by Mario Carneiro, 9-Jul-2013.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘𝐴) ∈ ℝ)
Theorem	rersqrtthlem 10958	Lemma for resqrtth 10959. (Contributed by Jim Kingdon, 10-Aug-2021.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (((√‘𝐴)↑2) = 𝐴 ∧ 0 ≤ (√‘𝐴)))
Theorem	resqrtth 10959	Square root theorem over the reals. Theorem I.35 of [Apostol] p. 29. (Contributed by Mario Carneiro, 9-Jul-2013.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ((√‘𝐴)↑2) = 𝐴)
Theorem	remsqsqrt 10960	Square of square root. (Contributed by Mario Carneiro, 10-Jul-2013.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ((√‘𝐴) · (√‘𝐴)) = 𝐴)
Theorem	sqrtge0 10961	The square root function is nonnegative for nonnegative input. (Contributed by NM, 26-May-1999.) (Revised by Mario Carneiro, 9-Jul-2013.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → 0 ≤ (√‘𝐴))
Theorem	sqrtgt0 10962	The square root function is positive for positive input. (Contributed by Mario Carneiro, 10-Jul-2013.) (Revised by Mario Carneiro, 6-Sep-2013.)
⊢ ((𝐴 ∈ ℝ ∧ 0 < 𝐴) → 0 < (√‘𝐴))
Theorem	sqrtmul 10963	Square root distributes over multiplication. (Contributed by NM, 30-Jul-1999.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → (√‘(𝐴 · 𝐵)) = ((√‘𝐴) · (√‘𝐵)))
Theorem	sqrtle 10964	Square root is monotonic. (Contributed by NM, 17-Mar-2005.) (Proof shortened by Mario Carneiro, 29-May-2016.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → (𝐴 ≤ 𝐵 ↔ (√‘𝐴) ≤ (√‘𝐵)))
Theorem	sqrtlt 10965	Square root is strictly monotonic. Closed form of sqrtlti 11065. (Contributed by Scott Fenton, 17-Apr-2014.) (Proof shortened by Mario Carneiro, 29-May-2016.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → (𝐴 < 𝐵 ↔ (√‘𝐴) < (√‘𝐵)))
Theorem	sqrt11ap 10966	Analogue to sqrt11 10967 but for apartness. (Contributed by Jim Kingdon, 11-Aug-2021.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((√‘𝐴) # (√‘𝐵) ↔ 𝐴 # 𝐵))
Theorem	sqrt11 10967	The square root function is one-to-one. Also see sqrt11ap 10966 which would follow easily from this given excluded middle, but which is proved another way without it. (Contributed by Scott Fenton, 11-Jun-2013.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((√‘𝐴) = (√‘𝐵) ↔ 𝐴 = 𝐵))
Theorem	sqrt00 10968	A square root is zero iff its argument is 0. (Contributed by NM, 27-Jul-1999.) (Proof shortened by Mario Carneiro, 29-May-2016.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ((√‘𝐴) = 0 ↔ 𝐴 = 0))
Theorem	rpsqrtcl 10969	The square root of a positive real is a positive real. (Contributed by NM, 22-Feb-2008.)
⊢ (𝐴 ∈ ℝ+ → (√‘𝐴) ∈ ℝ+)
Theorem	sqrtdiv 10970	Square root distributes over division. (Contributed by Mario Carneiro, 5-May-2016.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ 𝐵 ∈ ℝ+) → (√‘(𝐴 / 𝐵)) = ((√‘𝐴) / (√‘𝐵)))
Theorem	sqrtsq2 10971	Relationship between square root and squares. (Contributed by NM, 31-Jul-1999.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((√‘𝐴) = 𝐵 ↔ 𝐴 = (𝐵↑2)))
Theorem	sqrtsq 10972	Square root of square. (Contributed by NM, 14-Jan-2006.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘(𝐴↑2)) = 𝐴)
Theorem	sqrtmsq 10973	Square root of square. (Contributed by NM, 2-Aug-1999.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘(𝐴 · 𝐴)) = 𝐴)
Theorem	sqrt1 10974	The square root of 1 is 1. (Contributed by NM, 31-Jul-1999.)
⊢ (√‘1) = 1
Theorem	sqrt4 10975	The square root of 4 is 2. (Contributed by NM, 3-Aug-1999.)
⊢ (√‘4) = 2
Theorem	sqrt9 10976	The square root of 9 is 3. (Contributed by NM, 11-May-2004.)
⊢ (√‘9) = 3
Theorem	sqrt2gt1lt2 10977	The square root of 2 is bounded by 1 and 2. (Contributed by Roy F. Longton, 8-Aug-2005.) (Revised by Mario Carneiro, 6-Sep-2013.)
⊢ (1 < (√‘2) ∧ (√‘2) < 2)
Theorem	absneg 10978	Absolute value of negative. (Contributed by NM, 27-Feb-2005.)
⊢ (𝐴 ∈ ℂ → (abs‘-𝐴) = (abs‘𝐴))
Theorem	abscl 10979	Real closure of absolute value. (Contributed by NM, 3-Oct-1999.)
⊢ (𝐴 ∈ ℂ → (abs‘𝐴) ∈ ℝ)
Theorem	abscj 10980	The absolute value of a number and its conjugate are the same. Proposition 10-3.7(b) of [Gleason] p. 133. (Contributed by NM, 28-Apr-2005.)
⊢ (𝐴 ∈ ℂ → (abs‘(∗‘𝐴)) = (abs‘𝐴))
Theorem	absvalsq 10981	Square of value of absolute value function. (Contributed by NM, 16-Jan-2006.)
⊢ (𝐴 ∈ ℂ → ((abs‘𝐴)↑2) = (𝐴 · (∗‘𝐴)))
Theorem	absvalsq2 10982	Square of value of absolute value function. (Contributed by NM, 1-Feb-2007.)
⊢ (𝐴 ∈ ℂ → ((abs‘𝐴)↑2) = (((ℜ‘𝐴)↑2) + ((ℑ‘𝐴)↑2)))
Theorem	sqabsadd 10983	Square of absolute value of sum. Proposition 10-3.7(g) of [Gleason] p. 133. (Contributed by NM, 21-Jan-2007.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((abs‘(𝐴 + 𝐵))↑2) = ((((abs‘𝐴)↑2) + ((abs‘𝐵)↑2)) + (2 · (ℜ‘(𝐴 · (∗‘𝐵))))))
Theorem	sqabssub 10984	Square of absolute value of difference. (Contributed by NM, 21-Jan-2007.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((abs‘(𝐴 − 𝐵))↑2) = ((((abs‘𝐴)↑2) + ((abs‘𝐵)↑2)) − (2 · (ℜ‘(𝐴 · (∗‘𝐵))))))
Theorem	absval2 10985	Value of absolute value function. Definition 10.36 of [Gleason] p. 133. (Contributed by NM, 17-Mar-2005.)
⊢ (𝐴 ∈ ℂ → (abs‘𝐴) = (√‘(((ℜ‘𝐴)↑2) + ((ℑ‘𝐴)↑2))))
Theorem	abs0 10986	The absolute value of 0. (Contributed by NM, 26-Mar-2005.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ (abs‘0) = 0
Theorem	absi 10987	The absolute value of the imaginary unit. (Contributed by NM, 26-Mar-2005.)
⊢ (abs‘i) = 1
Theorem	absge0 10988	Absolute value is nonnegative. (Contributed by NM, 20-Nov-2004.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ (𝐴 ∈ ℂ → 0 ≤ (abs‘𝐴))
Theorem	absrpclap 10989	The absolute value of a number apart from zero is a positive real. (Contributed by Jim Kingdon, 11-Aug-2021.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐴 # 0) → (abs‘𝐴) ∈ ℝ+)
Theorem	abs00ap 10990	The absolute value of a number is apart from zero iff the number is apart from zero. (Contributed by Jim Kingdon, 11-Aug-2021.)
⊢ (𝐴 ∈ ℂ → ((abs‘𝐴) # 0 ↔ 𝐴 # 0))
Theorem	absext 10991	Strong extensionality for absolute value. (Contributed by Jim Kingdon, 12-Aug-2021.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((abs‘𝐴) # (abs‘𝐵) → 𝐴 # 𝐵))
Theorem	abs00 10992	The absolute value of a number is zero iff the number is zero. Also see abs00ap 10990 which is similar but for apartness. Proposition 10-3.7(c) of [Gleason] p. 133. (Contributed by NM, 26-Sep-2005.) (Proof shortened by Mario Carneiro, 29-May-2016.)
⊢ (𝐴 ∈ ℂ → ((abs‘𝐴) = 0 ↔ 𝐴 = 0))
Theorem	abs00ad 10993	A complex number is zero iff its absolute value is zero. Deduction form of abs00 10992. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → ((abs‘𝐴) = 0 ↔ 𝐴 = 0))
Theorem	abs00bd 10994	If a complex number is zero, its absolute value is zero. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 = 0)    ⇒   ⊢ (𝜑 → (abs‘𝐴) = 0)
Theorem	absreimsq 10995	Square of the absolute value of a number that has been decomposed into real and imaginary parts. (Contributed by NM, 1-Feb-2007.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → ((abs‘(𝐴 + (i · 𝐵)))↑2) = ((𝐴↑2) + (𝐵↑2)))
Theorem	absreim 10996	Absolute value of a number that has been decomposed into real and imaginary parts. (Contributed by NM, 14-Jan-2006.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (abs‘(𝐴 + (i · 𝐵))) = (√‘((𝐴↑2) + (𝐵↑2))))
Theorem	absmul 10997	Absolute value distributes over multiplication. Proposition 10-3.7(f) of [Gleason] p. 133. (Contributed by NM, 11-Oct-1999.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (abs‘(𝐴 · 𝐵)) = ((abs‘𝐴) · (abs‘𝐵)))
Theorem	absdivap 10998	Absolute value distributes over division. (Contributed by Jim Kingdon, 11-Aug-2021.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 # 0) → (abs‘(𝐴 / 𝐵)) = ((abs‘𝐴) / (abs‘𝐵)))
Theorem	absid 10999	A nonnegative number is its own absolute value. (Contributed by NM, 11-Oct-1999.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (abs‘𝐴) = 𝐴)
Theorem	abs1 11000	The absolute value of 1. Common special case. (Contributed by David A. Wheeler, 16-Jul-2016.)
⊢ (abs‘1) = 1