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	cjne0d 11501	A number which is nonzero has a complex conjugate which is nonzero. Also see cjap0d 11502 which is similar but for apartness. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 0)    ⇒   ⊢ (𝜑 → (∗‘𝐴) ≠ 0)
Theorem	cjap0d 11502	A number which is apart from zero has a complex conjugate which is apart from zero. (Contributed by Jim Kingdon, 11-Aug-2021.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 # 0)    ⇒   ⊢ (𝜑 → (∗‘𝐴) # 0)
Theorem	recjd 11503	Real part of a complex conjugate. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → (ℜ‘(∗‘𝐴)) = (ℜ‘𝐴))
Theorem	imcjd 11504	Imaginary part of a complex conjugate. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → (ℑ‘(∗‘𝐴)) = -(ℑ‘𝐴))
Theorem	cjmulrcld 11505	A complex number times its conjugate is real. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → (𝐴 · (∗‘𝐴)) ∈ ℝ)
Theorem	cjmulvald 11506	A complex number times its conjugate. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → (𝐴 · (∗‘𝐴)) = (((ℜ‘𝐴)↑2) + ((ℑ‘𝐴)↑2)))
Theorem	cjmulge0d 11507	A complex number times its conjugate is nonnegative. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → 0 ≤ (𝐴 · (∗‘𝐴)))
Theorem	renegd 11508	Real part of negative. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → (ℜ‘-𝐴) = -(ℜ‘𝐴))
Theorem	imnegd 11509	Imaginary part of negative. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → (ℑ‘-𝐴) = -(ℑ‘𝐴))
Theorem	cjnegd 11510	Complex conjugate of negative. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → (∗‘-𝐴) = -(∗‘𝐴))
Theorem	addcjd 11511	A number plus its conjugate is twice its real part. Compare Proposition 10-3.4(h) of [Gleason] p. 133. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → (𝐴 + (∗‘𝐴)) = (2 · (ℜ‘𝐴)))
Theorem	cjexpd 11512	Complex conjugate of positive integer exponentiation. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (∗‘(𝐴↑𝑁)) = ((∗‘𝐴)↑𝑁))
Theorem	readdd 11513	Real part distributes over addition. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → (ℜ‘(𝐴 + 𝐵)) = ((ℜ‘𝐴) + (ℜ‘𝐵)))
Theorem	imaddd 11514	Imaginary part distributes over addition. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → (ℑ‘(𝐴 + 𝐵)) = ((ℑ‘𝐴) + (ℑ‘𝐵)))
Theorem	resubd 11515	Real part distributes over subtraction. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → (ℜ‘(𝐴 − 𝐵)) = ((ℜ‘𝐴) − (ℜ‘𝐵)))
Theorem	imsubd 11516	Imaginary part distributes over subtraction. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → (ℑ‘(𝐴 − 𝐵)) = ((ℑ‘𝐴) − (ℑ‘𝐵)))
Theorem	remuld 11517	Real part of a product. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → (ℜ‘(𝐴 · 𝐵)) = (((ℜ‘𝐴) · (ℜ‘𝐵)) − ((ℑ‘𝐴) · (ℑ‘𝐵))))
Theorem	immuld 11518	Imaginary part of a product. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → (ℑ‘(𝐴 · 𝐵)) = (((ℜ‘𝐴) · (ℑ‘𝐵)) + ((ℑ‘𝐴) · (ℜ‘𝐵))))
Theorem	cjaddd 11519	Complex conjugate distributes over addition. Proposition 10-3.4(a) of [Gleason] p. 133. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → (∗‘(𝐴 + 𝐵)) = ((∗‘𝐴) + (∗‘𝐵)))
Theorem	cjmuld 11520	Complex conjugate distributes over multiplication. Proposition 10-3.4(c) of [Gleason] p. 133. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → (∗‘(𝐴 · 𝐵)) = ((∗‘𝐴) · (∗‘𝐵)))
Theorem	ipcnd 11521	Standard inner product on complex numbers. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → (ℜ‘(𝐴 · (∗‘𝐵))) = (((ℜ‘𝐴) · (ℜ‘𝐵)) + ((ℑ‘𝐴) · (ℑ‘𝐵))))
Theorem	cjdivapd 11522	Complex conjugate distributes over division. (Contributed by Jim Kingdon, 15-Jun-2020.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 # 0)    ⇒   ⊢ (𝜑 → (∗‘(𝐴 / 𝐵)) = ((∗‘𝐴) / (∗‘𝐵)))
Theorem	rered 11523	A real number equals its real part. One direction of Proposition 10-3.4(f) of [Gleason] p. 133. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℝ)    ⇒   ⊢ (𝜑 → (ℜ‘𝐴) = 𝐴)
Theorem	reim0d 11524	The imaginary part of a real number is 0. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℝ)    ⇒   ⊢ (𝜑 → (ℑ‘𝐴) = 0)
Theorem	cjred 11525	A real number equals its complex conjugate. Proposition 10-3.4(f) of [Gleason] p. 133. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℝ)    ⇒   ⊢ (𝜑 → (∗‘𝐴) = 𝐴)
Theorem	remul2d 11526	Real part of a product. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → (ℜ‘(𝐴 · 𝐵)) = (𝐴 · (ℜ‘𝐵)))
Theorem	immul2d 11527	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 11528	Real part of a division. Related to remul2 11427. (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 11529	Imaginary part of a division. Related to remul2 11427. (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 11530	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 11531	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 11532	Complex apartness in terms of real and imaginary parts. See also apreim 8776 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))))
4.8.3  Sequence convergence
Theorem	caucvgrelemrec 11533*	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 11534*	Lemma for caucvgre 11535. 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 11535*	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 11536	Lemma for cvg1n 11540. 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 11537*	Lemma for cvg1n 11540. 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 11538*	Lemma for cvg1n 11540. 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 11539*	Lemma for cvg1n 11540. 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 11540*	Convergence of real sequences. This is a version of caucvgre 11535 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 11541	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 11542*	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 11543*	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 11544*	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 11545*	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 11546*	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 11547*	A version of r19.2m 3579 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 11548	Lemma for recvguniq 11549. 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 11549*	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 11550	Extend class notation to include square root of a complex number.
class √
Syntax	cabs 11551	Extend class notation to include a function for the absolute value (modulus) of a complex number.
class abs
Definition	df-rsqrt 11552*	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 11553	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 11554*	Value of square root function. (Contributed by Jim Kingdon, 23-Aug-2020.)
⊢ (𝐴 ∈ ℝ → (√‘𝐴) = (℩𝑥 ∈ ℝ ((𝑥↑2) = 𝐴 ∧ 0 ≤ 𝑥)))
Theorem	absval 11555	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 11556	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 11557	Lemma for sqrt0 11558. (Contributed by Jim Kingdon, 26-Aug-2020.)
⊢ ((𝐴 ∈ ℝ ∧ ((𝐴↑2) = 0 ∧ 0 ≤ 𝐴)) ↔ 𝐴 = 0)
Theorem	sqrt0 11558	Square root of zero. (Contributed by Mario Carneiro, 9-Jul-2013.)
⊢ (√‘0) = 0
Theorem	resqrexlem1arp 11559	Lemma for resqrex 11580. 1 + 𝐴 is a positive real (expressed in a way that will help apply seqf 10719 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 11560*	Lemma for resqrex 11580. Applying the recursion rule yields a positive real (expressed in a way that will help apply seqf 10719 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 11561*	Lemma for resqrex 11580. 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 11562*	Lemma for resqrex 11580. 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 11563*	Lemma for resqrex 11580. 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 11564*	Lemma for resqrex 11580. 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 11565*	Lemma for resqrex 11580. 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 11566*	Lemma for resqrex 11580. 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 11567*	Lemma for resqrex 11580. 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 11568*	Lemma for resqrex 11580. 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 11569*	Lemma for resqrex 11580. 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 11570*	Lemma for resqrex 11580. 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 11571*	Lemma for resqrex 11580. 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 11572*	Lemma for resqrex 11580. 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 11573*	Lemma for resqrex 11580. 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 11574*	Lemma for resqrex 11580. 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 11575*	Lemma for resqrex 11580. 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 11576*	Lemma for resqrex 11580. 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 11577*	Lemma for resqrex 11580. 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 11578*	Lemma for resqrex 11580. 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 11579*	Lemma for resqrex 11580. 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 11580*	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 11581*	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 11582*	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 11583	Closure of the square root function. (Contributed by Mario Carneiro, 9-Jul-2013.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘𝐴) ∈ ℝ)
Theorem	rersqrtthlem 11584	Lemma for resqrtth 11585. (Contributed by Jim Kingdon, 10-Aug-2021.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (((√‘𝐴)↑2) = 𝐴 ∧ 0 ≤ (√‘𝐴)))
Theorem	resqrtth 11585	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 11586	Square of square root. (Contributed by Mario Carneiro, 10-Jul-2013.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ((√‘𝐴) · (√‘𝐴)) = 𝐴)
Theorem	sqrtge0 11587	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 11588	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 11589	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 11590	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 11591	Square root is strictly monotonic. Closed form of sqrtlti 11691. (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 11592	Analogue to sqrt11 11593 but for apartness. (Contributed by Jim Kingdon, 11-Aug-2021.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((√‘𝐴) # (√‘𝐵) ↔ 𝐴 # 𝐵))
Theorem	sqrt11 11593	The square root function is one-to-one. Also see sqrt11ap 11592 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 11594	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 11595	The square root of a positive real is a positive real. (Contributed by NM, 22-Feb-2008.)
⊢ (𝐴 ∈ ℝ+ → (√‘𝐴) ∈ ℝ+)
Theorem	sqrtdiv 11596	Square root distributes over division. (Contributed by Mario Carneiro, 5-May-2016.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ 𝐵 ∈ ℝ+) → (√‘(𝐴 / 𝐵)) = ((√‘𝐴) / (√‘𝐵)))
Theorem	sqrtsq2 11597	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 11598	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 11599	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 11600	The square root of 1 is 1. (Contributed by NM, 31-Jul-1999.)
⊢ (√‘1) = 1