P. 116 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 11501-11600   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	cvg1nlemcxze 11501	Lemma for cvg1n 11505. 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 11502*	Lemma for cvg1n 11505. 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 11503*	Lemma for cvg1n 11505. 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 11504*	Lemma for cvg1n 11505. 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 11505*	Convergence of real sequences. This is a version of caucvgre 11500 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 11506	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 11507*	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 11508*	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 11509*	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 11510*	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 11511*	A version of 19.29 1666 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 11512*	A version of r19.2m 3578 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 11513	Lemma for recvguniq 11514. 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 11514*	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)
4.8.4  Square root; absolute value
Syntax	csqrt 11515	Extend class notation to include square root of a complex number.
class √
Syntax	cabs 11516	Extend class notation to include a function for the absolute value (modulus) of a complex number.
class abs
Definition	df-rsqrt 11517*	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 11518	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 11519*	Value of square root function. (Contributed by Jim Kingdon, 23-Aug-2020.)
⊢ (𝐴 ∈ ℝ → (√‘𝐴) = (℩𝑥 ∈ ℝ ((𝑥↑2) = 𝐴 ∧ 0 ≤ 𝑥)))
Theorem	absval 11520	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 11521	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 11522	Lemma for sqrt0 11523. (Contributed by Jim Kingdon, 26-Aug-2020.)
⊢ ((𝐴 ∈ ℝ ∧ ((𝐴↑2) = 0 ∧ 0 ≤ 𝐴)) ↔ 𝐴 = 0)
Theorem	sqrt0 11523	Square root of zero. (Contributed by Mario Carneiro, 9-Jul-2013.)
⊢ (√‘0) = 0
Theorem	resqrexlem1arp 11524	Lemma for resqrex 11545. 1 + 𝐴 is a positive real (expressed in a way that will help apply seqf 10694 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 11525*	Lemma for resqrex 11545. Applying the recursion rule yields a positive real (expressed in a way that will help apply seqf 10694 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 11526*	Lemma for resqrex 11545. 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 11527*	Lemma for resqrex 11545. 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 11528*	Lemma for resqrex 11545. 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 11529*	Lemma for resqrex 11545. 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 11530*	Lemma for resqrex 11545. 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 11531*	Lemma for resqrex 11545. 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 11532*	Lemma for resqrex 11545. 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 11533*	Lemma for resqrex 11545. 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 11534*	Lemma for resqrex 11545. 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 11535*	Lemma for resqrex 11545. 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 11536*	Lemma for resqrex 11545. 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 11537*	Lemma for resqrex 11545. 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 11538*	Lemma for resqrex 11545. 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 11539*	Lemma for resqrex 11545. 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 11540*	Lemma for resqrex 11545. 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 11541*	Lemma for resqrex 11545. 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 11542*	Lemma for resqrex 11545. 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 11543*	Lemma for resqrex 11545. 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 11544*	Lemma for resqrex 11545. 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 11545*	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 11546*	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 11547*	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 11548	Closure of the square root function. (Contributed by Mario Carneiro, 9-Jul-2013.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘𝐴) ∈ ℝ)
Theorem	rersqrtthlem 11549	Lemma for resqrtth 11550. (Contributed by Jim Kingdon, 10-Aug-2021.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (((√‘𝐴)↑2) = 𝐴 ∧ 0 ≤ (√‘𝐴)))
Theorem	resqrtth 11550	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 11551	Square of square root. (Contributed by Mario Carneiro, 10-Jul-2013.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ((√‘𝐴) · (√‘𝐴)) = 𝐴)
Theorem	sqrtge0 11552	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 11553	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 11554	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 11555	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 11556	Square root is strictly monotonic. Closed form of sqrtlti 11656. (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 11557	Analogue to sqrt11 11558 but for apartness. (Contributed by Jim Kingdon, 11-Aug-2021.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((√‘𝐴) # (√‘𝐵) ↔ 𝐴 # 𝐵))
Theorem	sqrt11 11558	The square root function is one-to-one. Also see sqrt11ap 11557 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 11559	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 11560	The square root of a positive real is a positive real. (Contributed by NM, 22-Feb-2008.)
⊢ (𝐴 ∈ ℝ+ → (√‘𝐴) ∈ ℝ+)
Theorem	sqrtdiv 11561	Square root distributes over division. (Contributed by Mario Carneiro, 5-May-2016.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ 𝐵 ∈ ℝ+) → (√‘(𝐴 / 𝐵)) = ((√‘𝐴) / (√‘𝐵)))
Theorem	sqrtsq2 11562	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 11563	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 11564	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 11565	The square root of 1 is 1. (Contributed by NM, 31-Jul-1999.)
⊢ (√‘1) = 1
Theorem	sqrt4 11566	The square root of 4 is 2. (Contributed by NM, 3-Aug-1999.)
⊢ (√‘4) = 2
Theorem	sqrt9 11567	The square root of 9 is 3. (Contributed by NM, 11-May-2004.)
⊢ (√‘9) = 3
Theorem	sqrt2gt1lt2 11568	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 11569	Absolute value of negative. (Contributed by NM, 27-Feb-2005.)
⊢ (𝐴 ∈ ℂ → (abs‘-𝐴) = (abs‘𝐴))
Theorem	abscl 11570	Real closure of absolute value. (Contributed by NM, 3-Oct-1999.)
⊢ (𝐴 ∈ ℂ → (abs‘𝐴) ∈ ℝ)
Theorem	abscj 11571	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 11572	Square of value of absolute value function. (Contributed by NM, 16-Jan-2006.)
⊢ (𝐴 ∈ ℂ → ((abs‘𝐴)↑2) = (𝐴 · (∗‘𝐴)))
Theorem	absvalsq2 11573	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 11574	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 11575	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 11576	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 11577	The absolute value of 0. (Contributed by NM, 26-Mar-2005.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ (abs‘0) = 0
Theorem	absi 11578	The absolute value of the imaginary unit. (Contributed by NM, 26-Mar-2005.)
⊢ (abs‘i) = 1
Theorem	absge0 11579	Absolute value is nonnegative. (Contributed by NM, 20-Nov-2004.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ (𝐴 ∈ ℂ → 0 ≤ (abs‘𝐴))
Theorem	absrpclap 11580	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 11581	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 11582	Strong extensionality for absolute value. (Contributed by Jim Kingdon, 12-Aug-2021.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((abs‘𝐴) # (abs‘𝐵) → 𝐴 # 𝐵))
Theorem	abs00 11583	The absolute value of a number is zero iff the number is zero. Also see abs00ap 11581 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 11584	A complex number is zero iff its absolute value is zero. Deduction form of abs00 11583. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → ((abs‘𝐴) = 0 ↔ 𝐴 = 0))
Theorem	abs00bd 11585	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 11586	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 11587	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 11588	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 11589	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 11590	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 11591	The absolute value of 1. Common special case. (Contributed by David A. Wheeler, 16-Jul-2016.)
⊢ (abs‘1) = 1
Theorem	absnid 11592	A negative number is the negative of its own absolute value. (Contributed by NM, 27-Feb-2005.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐴 ≤ 0) → (abs‘𝐴) = -𝐴)
Theorem	leabs 11593	A real number is less than or equal to its absolute value. (Contributed by NM, 27-Feb-2005.)
⊢ (𝐴 ∈ ℝ → 𝐴 ≤ (abs‘𝐴))
Theorem	qabsor 11594	The absolute value of a rational number is either that number or its negative. (Contributed by Jim Kingdon, 8-Nov-2021.)
⊢ (𝐴 ∈ ℚ → ((abs‘𝐴) = 𝐴 ∨ (abs‘𝐴) = -𝐴))
Theorem	qabsord 11595	The absolute value of a rational number is either that number or its negative. (Contributed by Jim Kingdon, 8-Nov-2021.)
⊢ (𝜑 → 𝐴 ∈ ℚ)    ⇒   ⊢ (𝜑 → ((abs‘𝐴) = 𝐴 ∨ (abs‘𝐴) = -𝐴))
Theorem	absre 11596	Absolute value of a real number. (Contributed by NM, 17-Mar-2005.)
⊢ (𝐴 ∈ ℝ → (abs‘𝐴) = (√‘(𝐴↑2)))
Theorem	absresq 11597	Square of the absolute value of a real number. (Contributed by NM, 16-Jan-2006.)
⊢ (𝐴 ∈ ℝ → ((abs‘𝐴)↑2) = (𝐴↑2))
Theorem	absexp 11598	Absolute value of positive integer exponentiation. (Contributed by NM, 5-Jan-2006.)
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (abs‘(𝐴↑𝑁)) = ((abs‘𝐴)↑𝑁))
Theorem	absexpzap 11599	Absolute value of integer exponentiation. (Contributed by Jim Kingdon, 11-Aug-2021.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐴 # 0 ∧ 𝑁 ∈ ℤ) → (abs‘(𝐴↑𝑁)) = ((abs‘𝐴)↑𝑁))
Theorem	abssq 11600	Square can be moved in and out of absolute value. (Contributed by Scott Fenton, 18-Apr-2014.) (Proof shortened by Mario Carneiro, 29-May-2016.)
⊢ (𝐴 ∈ ℂ → ((abs‘𝐴)↑2) = (abs‘(𝐴↑2)))