P. 107 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 10601-10700   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	rexfiuz 10601*	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 10602*	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 10603*	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 10604*	A version of 19.29 1567 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 10605*	A version of r19.2m 3396 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 10606	Lemma for recvguniq 10607. 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 10607*	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)
3.7.4  Square root; absolute value
Syntax	csqrt 10608	Extend class notation to include square root of a complex number.
class √
Syntax	cabs 10609	Extend class notation to include a function for the absolute value (modulus) of a complex number.
class abs
Definition	df-rsqrt 10610*	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 10611	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 10612*	Value of square root function. (Contributed by Jim Kingdon, 23-Aug-2020.)
⊢ (𝐴 ∈ ℝ → (√‘𝐴) = (℩𝑥 ∈ ℝ ((𝑥↑2) = 𝐴 ∧ 0 ≤ 𝑥)))
Theorem	absval 10613	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 10614	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 10615	Lemma for sqrt0 10616. (Contributed by Jim Kingdon, 26-Aug-2020.)
⊢ ((𝐴 ∈ ℝ ∧ ((𝐴↑2) = 0 ∧ 0 ≤ 𝐴)) ↔ 𝐴 = 0)
Theorem	sqrt0 10616	Square root of zero. (Contributed by Mario Carneiro, 9-Jul-2013.)
⊢ (√‘0) = 0
Theorem	resqrexlem1arp 10617	Lemma for resqrex 10638. 1 + 𝐴 is a positive real (expressed in a way that will help apply seqf 10075 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 10618*	Lemma for resqrex 10638. Applying the recursion rule yields a positive real (expressed in a way that will help apply seqf 10075 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 10619*	Lemma for resqrex 10638. 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 10620*	Lemma for resqrex 10638. 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 10621*	Lemma for resqrex 10638. 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 10622*	Lemma for resqrex 10638. 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 10623*	Lemma for resqrex 10638. 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 10624*	Lemma for resqrex 10638. 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 10625*	Lemma for resqrex 10638. 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 10626*	Lemma for resqrex 10638. 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 10627*	Lemma for resqrex 10638. 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 10628*	Lemma for resqrex 10638. 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 10629*	Lemma for resqrex 10638. 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 10630*	Lemma for resqrex 10638. 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 10631*	Lemma for resqrex 10638. 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 10632*	Lemma for resqrex 10638. 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 10633*	Lemma for resqrex 10638. 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 10634*	Lemma for resqrex 10638. 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 10635*	Lemma for resqrex 10638. 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 10636*	Lemma for resqrex 10638. 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 10637*	Lemma for resqrex 10638. 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 10638*	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 10639*	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 10640*	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 10641	Closure of the square root function. (Contributed by Mario Carneiro, 9-Jul-2013.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘𝐴) ∈ ℝ)
Theorem	rersqrtthlem 10642	Lemma for resqrtth 10643. (Contributed by Jim Kingdon, 10-Aug-2021.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (((√‘𝐴)↑2) = 𝐴 ∧ 0 ≤ (√‘𝐴)))
Theorem	resqrtth 10643	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 10644	Square of square root. (Contributed by Mario Carneiro, 10-Jul-2013.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ((√‘𝐴) · (√‘𝐴)) = 𝐴)
Theorem	sqrtge0 10645	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 10646	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 10647	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 10648	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 10649	Square root is strictly monotonic. Closed form of sqrtlti 10749. (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 10650	Analogue to sqrt11 10651 but for apartness. (Contributed by Jim Kingdon, 11-Aug-2021.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((√‘𝐴) # (√‘𝐵) ↔ 𝐴 # 𝐵))
Theorem	sqrt11 10651	The square root function is one-to-one. Also see sqrt11ap 10650 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 10652	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 10653	The square root of a positive real is a positive real. (Contributed by NM, 22-Feb-2008.)
⊢ (𝐴 ∈ ℝ+ → (√‘𝐴) ∈ ℝ+)
Theorem	sqrtdiv 10654	Square root distributes over division. (Contributed by Mario Carneiro, 5-May-2016.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ 𝐵 ∈ ℝ+) → (√‘(𝐴 / 𝐵)) = ((√‘𝐴) / (√‘𝐵)))
Theorem	sqrtsq2 10655	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 10656	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 10657	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 10658	The square root of 1 is 1. (Contributed by NM, 31-Jul-1999.)
⊢ (√‘1) = 1
Theorem	sqrt4 10659	The square root of 4 is 2. (Contributed by NM, 3-Aug-1999.)
⊢ (√‘4) = 2
Theorem	sqrt9 10660	The square root of 9 is 3. (Contributed by NM, 11-May-2004.)
⊢ (√‘9) = 3
Theorem	sqrt2gt1lt2 10661	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 10662	Absolute value of negative. (Contributed by NM, 27-Feb-2005.)
⊢ (𝐴 ∈ ℂ → (abs‘-𝐴) = (abs‘𝐴))
Theorem	abscl 10663	Real closure of absolute value. (Contributed by NM, 3-Oct-1999.)
⊢ (𝐴 ∈ ℂ → (abs‘𝐴) ∈ ℝ)
Theorem	abscj 10664	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 10665	Square of value of absolute value function. (Contributed by NM, 16-Jan-2006.)
⊢ (𝐴 ∈ ℂ → ((abs‘𝐴)↑2) = (𝐴 · (∗‘𝐴)))
Theorem	absvalsq2 10666	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 10667	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 10668	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 10669	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 10670	The absolute value of 0. (Contributed by NM, 26-Mar-2005.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ (abs‘0) = 0
Theorem	absi 10671	The absolute value of the imaginary unit. (Contributed by NM, 26-Mar-2005.)
⊢ (abs‘i) = 1
Theorem	absge0 10672	Absolute value is nonnegative. (Contributed by NM, 20-Nov-2004.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ (𝐴 ∈ ℂ → 0 ≤ (abs‘𝐴))
Theorem	absrpclap 10673	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 10674	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 10675	Strong extensionality for absolute value. (Contributed by Jim Kingdon, 12-Aug-2021.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((abs‘𝐴) # (abs‘𝐵) → 𝐴 # 𝐵))
Theorem	abs00 10676	The absolute value of a number is zero iff the number is zero. Also see abs00ap 10674 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 10677	A complex number is zero iff its absolute value is zero. Deduction form of abs00 10676. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → ((abs‘𝐴) = 0 ↔ 𝐴 = 0))
Theorem	abs00bd 10678	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 10679	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 10680	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 10681	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 10682	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 10683	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 10684	The absolute value of 1. Common special case. (Contributed by David A. Wheeler, 16-Jul-2016.)
⊢ (abs‘1) = 1
Theorem	absnid 10685	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 10686	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 10687	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 10688	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 10689	Absolute value of a real number. (Contributed by NM, 17-Mar-2005.)
⊢ (𝐴 ∈ ℝ → (abs‘𝐴) = (√‘(𝐴↑2)))
Theorem	absresq 10690	Square of the absolute value of a real number. (Contributed by NM, 16-Jan-2006.)
⊢ (𝐴 ∈ ℝ → ((abs‘𝐴)↑2) = (𝐴↑2))
Theorem	absexp 10691	Absolute value of positive integer exponentiation. (Contributed by NM, 5-Jan-2006.)
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (abs‘(𝐴↑𝑁)) = ((abs‘𝐴)↑𝑁))
Theorem	absexpzap 10692	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 10693	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)))
Theorem	sqabs 10694	The squares of two reals are equal iff their absolute values are equal. (Contributed by NM, 6-Mar-2009.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → ((𝐴↑2) = (𝐵↑2) ↔ (abs‘𝐴) = (abs‘𝐵)))
Theorem	absrele 10695	The absolute value of a complex number is greater than or equal to the absolute value of its real part. (Contributed by NM, 1-Apr-2005.)
⊢ (𝐴 ∈ ℂ → (abs‘(ℜ‘𝐴)) ≤ (abs‘𝐴))
Theorem	absimle 10696	The absolute value of a complex number is greater than or equal to the absolute value of its imaginary part. (Contributed by NM, 17-Mar-2005.) (Proof shortened by Mario Carneiro, 29-May-2016.)
⊢ (𝐴 ∈ ℂ → (abs‘(ℑ‘𝐴)) ≤ (abs‘𝐴))
Theorem	nn0abscl 10697	The absolute value of an integer is a nonnegative integer. (Contributed by NM, 27-Feb-2005.)
⊢ (𝐴 ∈ ℤ → (abs‘𝐴) ∈ ℕ0)
Theorem	zabscl 10698	The absolute value of an integer is an integer. (Contributed by Stefan O'Rear, 24-Sep-2014.)
⊢ (𝐴 ∈ ℤ → (abs‘𝐴) ∈ ℤ)
Theorem	ltabs 10699	A number which is less than its absolute value is negative. (Contributed by Jim Kingdon, 12-Aug-2021.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐴 < (abs‘𝐴)) → 𝐴 < 0)
Theorem	abslt 10700	Absolute value and 'less than' relation. (Contributed by NM, 6-Apr-2005.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → ((abs‘𝐴) < 𝐵 ↔ (-𝐵 < 𝐴 ∧ 𝐴 < 𝐵)))