P. 126 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 12501-12600   *Has distinct variable group(s)

Type	Label	Description
Statement
5.2.11  Fundamental theorem of arithmetic
Theorem	1arithlem1 12501*	Lemma for 1arith 12505. (Contributed by Mario Carneiro, 30-May-2014.)
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    ⇒   ⊢ (𝑁 ∈ ℕ → (𝑀‘𝑁) = (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑁)))
Theorem	1arithlem2 12502*	Lemma for 1arith 12505. (Contributed by Mario Carneiro, 30-May-2014.)
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    ⇒   ⊢ ((𝑁 ∈ ℕ ∧ 𝑃 ∈ ℙ) → ((𝑀‘𝑁)‘𝑃) = (𝑃 pCnt 𝑁))
Theorem	1arithlem3 12503*	Lemma for 1arith 12505. (Contributed by Mario Carneiro, 30-May-2014.)
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    ⇒   ⊢ (𝑁 ∈ ℕ → (𝑀‘𝑁):ℙ⟶ℕ0)
Theorem	1arithlem4 12504*	Lemma for 1arith 12505. (Contributed by Mario Carneiro, 30-May-2014.)
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    &   ⊢ 𝐺 = (𝑦 ∈ ℕ ↦ if(𝑦 ∈ ℙ, (𝑦↑(𝐹‘𝑦)), 1))    &   ⊢ (𝜑 → 𝐹:ℙ⟶ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ ((𝜑 ∧ (𝑞 ∈ ℙ ∧ 𝑁 ≤ 𝑞)) → (𝐹‘𝑞) = 0)    ⇒   ⊢ (𝜑 → ∃𝑥 ∈ ℕ 𝐹 = (𝑀‘𝑥))
Theorem	1arith 12505*	Fundamental theorem of arithmetic, where a prime factorization is represented as a sequence of prime exponents, for which only finitely many primes have nonzero exponent. The function 𝑀 maps the set of positive integers one-to-one onto the set of prime factorizations 𝑅. (Contributed by Paul Chapman, 17-Nov-2012.) (Proof shortened by Mario Carneiro, 30-May-2014.)
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    &   ⊢ 𝑅 = {𝑒 ∈ (ℕ0 ↑𝑚 ℙ) ∣ (◡𝑒 “ ℕ) ∈ Fin}    ⇒   ⊢ 𝑀:ℕ–1-1-onto→𝑅
Theorem	1arith2 12506*	Fundamental theorem of arithmetic, where a prime factorization is represented as a finite monotonic 1-based sequence of primes. Every positive integer has a unique prime factorization. Theorem 1.10 in [ApostolNT] p. 17. This is Metamath 100 proof #80. (Contributed by Paul Chapman, 17-Nov-2012.) (Revised by Mario Carneiro, 30-May-2014.)
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    &   ⊢ 𝑅 = {𝑒 ∈ (ℕ0 ↑𝑚 ℙ) ∣ (◡𝑒 “ ℕ) ∈ Fin}    ⇒   ⊢ ∀𝑧 ∈ ℕ ∃!𝑔 ∈ 𝑅 (𝑀‘𝑧) = 𝑔
5.2.12  Lagrange's four-square theorem
Syntax	cgz 12507	Extend class notation with the set of gaussian integers.
class ℤ[i]
Definition	df-gz 12508	Define the set of gaussian integers, which are complex numbers whose real and imaginary parts are integers. (Note that the [i] is actually part of the symbol token and has no independent meaning.) (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ ℤ[i] = {𝑥 ∈ ℂ ∣ ((ℜ‘𝑥) ∈ ℤ ∧ (ℑ‘𝑥) ∈ ℤ)}
Theorem	elgz 12509	Elementhood in the gaussian integers. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] ↔ (𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ))
Theorem	gzcn 12510	A gaussian integer is a complex number. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → 𝐴 ∈ ℂ)
Theorem	zgz 12511	An integer is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ → 𝐴 ∈ ℤ[i])
Theorem	igz 12512	i is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ i ∈ ℤ[i]
Theorem	gznegcl 12513	The gaussian integers are closed under negation. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → -𝐴 ∈ ℤ[i])
Theorem	gzcjcl 12514	The gaussian integers are closed under conjugation. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → (∗‘𝐴) ∈ ℤ[i])
Theorem	gzaddcl 12515	The gaussian integers are closed under addition. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ ((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴 + 𝐵) ∈ ℤ[i])
Theorem	gzmulcl 12516	The gaussian integers are closed under multiplication. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ ((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴 · 𝐵) ∈ ℤ[i])
Theorem	gzreim 12517	Construct a gaussian integer from real and imaginary parts. (Contributed by Mario Carneiro, 16-Jul-2014.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → (𝐴 + (i · 𝐵)) ∈ ℤ[i])
Theorem	gzsubcl 12518	The gaussian integers are closed under subtraction. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ ((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴 − 𝐵) ∈ ℤ[i])
Theorem	gzabssqcl 12519	The squared norm of a gaussian integer is an integer. (Contributed by Mario Carneiro, 16-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → ((abs‘𝐴)↑2) ∈ ℕ0)
Theorem	4sqlem5 12520	Lemma for 4sq 12548. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    ⇒   ⊢ (𝜑 → (𝐵 ∈ ℤ ∧ ((𝐴 − 𝐵) / 𝑀) ∈ ℤ))
Theorem	4sqlem6 12521	Lemma for 4sq 12548. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    ⇒   ⊢ (𝜑 → (-(𝑀 / 2) ≤ 𝐵 ∧ 𝐵 < (𝑀 / 2)))
Theorem	4sqlem7 12522	Lemma for 4sq 12548. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    ⇒   ⊢ (𝜑 → (𝐵↑2) ≤ (((𝑀↑2) / 2) / 2))
Theorem	4sqlem8 12523	Lemma for 4sq 12548. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    ⇒   ⊢ (𝜑 → 𝑀 ∥ ((𝐴↑2) − (𝐵↑2)))
Theorem	4sqlem9 12524	Lemma for 4sq 12548. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ ((𝜑 ∧ 𝜓) → (𝐵↑2) = 0)    ⇒   ⊢ ((𝜑 ∧ 𝜓) → (𝑀↑2) ∥ (𝐴↑2))
Theorem	4sqlem10 12525	Lemma for 4sq 12548. (Contributed by Mario Carneiro, 16-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ ((𝜑 ∧ 𝜓) → ((((𝑀↑2) / 2) / 2) − (𝐵↑2)) = 0)    ⇒   ⊢ ((𝜑 ∧ 𝜓) → (𝑀↑2) ∥ ((𝐴↑2) − (((𝑀↑2) / 2) / 2)))
Theorem	4sqlem1 12526*	Lemma for 4sq 12548. The set 𝑆 is the set of all numbers that are expressible as a sum of four squares. Our goal is to show that 𝑆 = ℕ0; here we show one subset direction. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ 𝑆 ⊆ ℕ0
Theorem	4sqlem2 12527*	Lemma for 4sq 12548. Change bound variables in 𝑆. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ (𝐴 ∈ 𝑆 ↔ ∃𝑎 ∈ ℤ ∃𝑏 ∈ ℤ ∃𝑐 ∈ ℤ ∃𝑑 ∈ ℤ 𝐴 = (((𝑎↑2) + (𝑏↑2)) + ((𝑐↑2) + (𝑑↑2))))
Theorem	4sqlem3 12528*	Lemma for 4sq 12548. Sufficient condition to be in 𝑆. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝐶 ∈ ℤ ∧ 𝐷 ∈ ℤ)) → (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))) ∈ 𝑆)
Theorem	4sqlem4a 12529*	Lemma for 4sqlem4 12530. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ ((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (((abs‘𝐴)↑2) + ((abs‘𝐵)↑2)) ∈ 𝑆)
Theorem	4sqlem4 12530*	Lemma for 4sq 12548. We can express the four-square property more compactly in terms of gaussian integers, because the norms of gaussian integers are exactly sums of two squares. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ (𝐴 ∈ 𝑆 ↔ ∃𝑢 ∈ ℤ[i] ∃𝑣 ∈ ℤ[i] 𝐴 = (((abs‘𝑢)↑2) + ((abs‘𝑣)↑2)))
Theorem	mul4sqlem 12531*	Lemma for mul4sq 12532: algebraic manipulations. The extra assumptions involving 𝑀 would let us know not just that the product is a sum of squares, but also that it preserves divisibility by 𝑀. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝐴 ∈ ℤ[i])    &   ⊢ (𝜑 → 𝐵 ∈ ℤ[i])    &   ⊢ (𝜑 → 𝐶 ∈ ℤ[i])    &   ⊢ (𝜑 → 𝐷 ∈ ℤ[i])    &   ⊢ 𝑋 = (((abs‘𝐴)↑2) + ((abs‘𝐵)↑2))    &   ⊢ 𝑌 = (((abs‘𝐶)↑2) + ((abs‘𝐷)↑2))    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → ((𝐴 − 𝐶) / 𝑀) ∈ ℤ[i])    &   ⊢ (𝜑 → ((𝐵 − 𝐷) / 𝑀) ∈ ℤ[i])    &   ⊢ (𝜑 → (𝑋 / 𝑀) ∈ ℕ0)    ⇒   ⊢ (𝜑 → ((𝑋 / 𝑀) · (𝑌 / 𝑀)) ∈ 𝑆)
Theorem	mul4sq 12532*	Euler's four-square identity: The product of two sums of four squares is also a sum of four squares. This is usually quoted as an explicit formula involving eight real variables; we save some time by working with complex numbers (gaussian integers) instead, so that we only have to work with four variables, and also hiding the actual formula for the product in the proof of mul4sqlem 12531. (For the curious, the explicit formula that is used is ( ∣ 𝑎 ∣ ↑2 + ∣ 𝑏 ∣ ↑2)( ∣ 𝑐 ∣ ↑2 + ∣ 𝑑 ∣ ↑2) = ∣ 𝑎∗ · 𝑐 + 𝑏 · 𝑑∗ ∣ ↑2 + ∣ 𝑎∗ · 𝑑 − 𝑏 · 𝑐∗ ∣ ↑2.) (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ ((𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆) → (𝐴 · 𝐵) ∈ 𝑆)
Theorem	4sqlemafi 12533*	Lemma for 4sq 12548. 𝐴 is finite. (Contributed by Jim Kingdon, 24-May-2025.)
⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    ⇒   ⊢ (𝜑 → 𝐴 ∈ Fin)
Theorem	4sqlemffi 12534*	Lemma for 4sq 12548. ran 𝐹 is finite. (Contributed by Jim Kingdon, 24-May-2025.)
⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   ⊢ 𝐹 = (𝑣 ∈ 𝐴 ↦ ((𝑃 − 1) − 𝑣))    ⇒   ⊢ (𝜑 → ran 𝐹 ∈ Fin)
Theorem	4sqleminfi 12535*	Lemma for 4sq 12548. 𝐴 ∩ ran 𝐹 is finite. (Contributed by Jim Kingdon, 24-May-2025.)
⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   ⊢ 𝐹 = (𝑣 ∈ 𝐴 ↦ ((𝑃 − 1) − 𝑣))    ⇒   ⊢ (𝜑 → (𝐴 ∩ ran 𝐹) ∈ Fin)
Theorem	4sqexercise1 12536*	Exercise which may help in understanding the proof of 4sqlemsdc 12538. (Contributed by Jim Kingdon, 25-May-2025.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ 𝑛 = (𝑥↑2)}    ⇒   ⊢ (𝐴 ∈ ℕ0 → DECID 𝐴 ∈ 𝑆)
Theorem	4sqexercise2 12537*	Exercise which may help in understanding the proof of 4sqlemsdc 12538. (Contributed by Jim Kingdon, 30-May-2025.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑛 = ((𝑥↑2) + (𝑦↑2))}    ⇒   ⊢ (𝐴 ∈ ℕ0 → DECID 𝐴 ∈ 𝑆)
Theorem	4sqlemsdc 12538*	Lemma for 4sq 12548. The property of being the sum of four squares is decidable. The proof involves showing that (for a particular 𝐴) there are only a finite number of possible ways that it could be the sum of four squares, so checking each of those possibilities in turn decides whether the number is the sum of four squares. If this proof is hard to follow, especially because of its length, the simplified versions at 4sqexercise1 12536 and 4sqexercise2 12537 may help clarify, as they are using very much the same techniques on simplified versions of this lemma. (Contributed by Jim Kingdon, 25-May-2025.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ (𝐴 ∈ ℕ0 → DECID 𝐴 ∈ 𝑆)
Theorem	4sqlem11 12539*	Lemma for 4sq 12548. Use the pigeonhole principle to show that the sets {𝑚↑2 ∣ 𝑚 ∈ (0...𝑁)} and {-1 − 𝑛↑2 ∣ 𝑛 ∈ (0...𝑁)} have a common element, mod 𝑃. Note that although the conclusion is stated in terms of 𝐴 ∩ ran 𝐹 being nonempty, it is also inhabited by 4sqleminfi 12535 and fin0 6941. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   ⊢ 𝐹 = (𝑣 ∈ 𝐴 ↦ ((𝑃 − 1) − 𝑣))    ⇒   ⊢ (𝜑 → (𝐴 ∩ ran 𝐹) ≠ ∅)
Theorem	4sqlem12 12540*	Lemma for 4sq 12548. For any odd prime 𝑃, there is a 𝑘 < 𝑃 such that 𝑘𝑃 − 1 is a sum of two squares. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   ⊢ 𝐹 = (𝑣 ∈ 𝐴 ↦ ((𝑃 − 1) − 𝑣))    ⇒   ⊢ (𝜑 → ∃𝑘 ∈ (1...(𝑃 − 1))∃𝑢 ∈ ℤ[i] (((abs‘𝑢)↑2) + 1) = (𝑘 · 𝑃))
Theorem	4sqlem13m 12541*	Lemma for 4sq 12548. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    ⇒   ⊢ (𝜑 → (∃𝑗 𝑗 ∈ 𝑇 ∧ 𝑀 < 𝑃))
Theorem	4sqlem14 12542*	Lemma for 4sq 12548. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ 𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   ⊢ (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))    ⇒   ⊢ (𝜑 → 𝑅 ∈ ℕ0)
Theorem	4sqlem15 12543*	Lemma for 4sq 12548. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ 𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   ⊢ (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))    ⇒   ⊢ ((𝜑 ∧ 𝑅 = 𝑀) → ((((((𝑀↑2) / 2) / 2) − (𝐸↑2)) = 0 ∧ ((((𝑀↑2) / 2) / 2) − (𝐹↑2)) = 0) ∧ (((((𝑀↑2) / 2) / 2) − (𝐺↑2)) = 0 ∧ ((((𝑀↑2) / 2) / 2) − (𝐻↑2)) = 0)))
Theorem	4sqlem16 12544*	Lemma for 4sq 12548. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ 𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   ⊢ (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))    ⇒   ⊢ (𝜑 → (𝑅 ≤ 𝑀 ∧ ((𝑅 = 0 ∨ 𝑅 = 𝑀) → (𝑀↑2) ∥ (𝑀 · 𝑃))))
Theorem	4sqlem17 12545*	Lemma for 4sq 12548. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ 𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   ⊢ (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))    ⇒   ⊢ ¬ 𝜑
Theorem	4sqlem18 12546*	Lemma for 4sq 12548. Inductive step, odd prime case. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    ⇒   ⊢ (𝜑 → 𝑃 ∈ 𝑆)
Theorem	4sqlem19 12547*	Lemma for 4sq 12548. The proof is by strong induction - we show that if all the integers less than 𝑘 are in 𝑆, then 𝑘 is as well. In this part of the proof we do the induction argument and dispense with all the cases except the odd prime case, which is sent to 4sqlem18 12546. If 𝑘 is 0, 1, 2, we show 𝑘 ∈ 𝑆 directly; otherwise if 𝑘 is composite, 𝑘 is the product of two numbers less than it (and hence in 𝑆 by assumption), so by mul4sq 12532 𝑘 ∈ 𝑆. (Contributed by Mario Carneiro, 14-Jul-2014.) (Revised by Mario Carneiro, 20-Jun-2015.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ ℕ0 = 𝑆
Theorem	4sq 12548*	Lagrange's four-square theorem, or Bachet's conjecture: every nonnegative integer is expressible as a sum of four squares. This is Metamath 100 proof #19. (Contributed by Mario Carneiro, 16-Jul-2014.)
⊢ (𝐴 ∈ ℕ0 ↔ ∃𝑎 ∈ ℤ ∃𝑏 ∈ ℤ ∃𝑐 ∈ ℤ ∃𝑑 ∈ ℤ 𝐴 = (((𝑎↑2) + (𝑏↑2)) + ((𝑐↑2) + (𝑑↑2))))
5.3  Cardinality of real and complex number subsets
5.3.1  Countability of integers and rationals
Theorem	oddennn 12549	There are as many odd positive integers as there are positive integers. (Contributed by Jim Kingdon, 11-May-2022.)
⊢ {𝑧 ∈ ℕ ∣ ¬ 2 ∥ 𝑧} ≈ ℕ
Theorem	evenennn 12550	There are as many even positive integers as there are positive integers. (Contributed by Jim Kingdon, 12-May-2022.)
⊢ {𝑧 ∈ ℕ ∣ 2 ∥ 𝑧} ≈ ℕ
Theorem	xpnnen 12551	The Cartesian product of the set of positive integers with itself is equinumerous to the set of positive integers. (Contributed by NM, 1-Aug-2004.)
⊢ (ℕ × ℕ) ≈ ℕ
Theorem	xpomen 12552	The Cartesian product of omega (the set of ordinal natural numbers) with itself is equinumerous to omega. Exercise 1 of [Enderton] p. 133. (Contributed by NM, 23-Jul-2004.)
⊢ (ω × ω) ≈ ω
Theorem	xpct 12553	The cartesian product of two sets dominated by ω is dominated by ω. (Contributed by Thierry Arnoux, 24-Sep-2017.)
⊢ ((𝐴 ≼ ω ∧ 𝐵 ≼ ω) → (𝐴 × 𝐵) ≼ ω)
Theorem	unennn 12554	The union of two disjoint countably infinite sets is countably infinite. (Contributed by Jim Kingdon, 13-May-2022.)
⊢ ((𝐴 ≈ ℕ ∧ 𝐵 ≈ ℕ ∧ (𝐴 ∩ 𝐵) = ∅) → (𝐴 ∪ 𝐵) ≈ ℕ)
Theorem	znnen 12555	The set of integers and the set of positive integers are equinumerous. Corollary 8.1.23 of [AczelRathjen], p. 75. (Contributed by NM, 31-Jul-2004.)
⊢ ℤ ≈ ℕ
Theorem	ennnfonelemdc 12556*	Lemma for ennnfone 12582. A direct consequence of fidcenumlemrk 7013. (Contributed by Jim Kingdon, 15-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → 𝑃 ∈ ω)    ⇒   ⊢ (𝜑 → DECID (𝐹‘𝑃) ∈ (𝐹 “ 𝑃))
Theorem	ennnfonelemk 12557*	Lemma for ennnfone 12582. (Contributed by Jim Kingdon, 15-Jul-2023.)
⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → 𝐾 ∈ ω)    &   ⊢ (𝜑 → 𝑁 ∈ ω)    &   ⊢ (𝜑 → ∀𝑗 ∈ suc 𝑁(𝐹‘𝐾) ≠ (𝐹‘𝑗))    ⇒   ⊢ (𝜑 → 𝑁 ∈ 𝐾)
Theorem	ennnfonelemj0 12558*	Lemma for ennnfone 12582. Initial state for 𝐽. (Contributed by Jim Kingdon, 20-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    ⇒   ⊢ (𝜑 → (𝐽‘0) ∈ {𝑔 ∈ (𝐴 ↑pm ω) ∣ dom 𝑔 ∈ ω})
Theorem	ennnfonelemjn 12559*	Lemma for ennnfone 12582. Non-initial state for 𝐽. (Contributed by Jim Kingdon, 20-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    ⇒   ⊢ ((𝜑 ∧ 𝑓 ∈ (ℤ≥‘(0 + 1))) → (𝐽‘𝑓) ∈ ω)
Theorem	ennnfonelemg 12560*	Lemma for ennnfone 12582. Closure for 𝐺. (Contributed by Jim Kingdon, 20-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    ⇒   ⊢ ((𝜑 ∧ (𝑓 ∈ {𝑔 ∈ (𝐴 ↑pm ω) ∣ dom 𝑔 ∈ ω} ∧ 𝑗 ∈ ω)) → (𝑓𝐺𝑗) ∈ {𝑔 ∈ (𝐴 ↑pm ω) ∣ dom 𝑔 ∈ ω})
Theorem	ennnfonelemh 12561*	Lemma for ennnfone 12582. (Contributed by Jim Kingdon, 8-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    ⇒   ⊢ (𝜑 → 𝐻:ℕ0⟶(𝐴 ↑pm ω))
Theorem	ennnfonelem0 12562*	Lemma for ennnfone 12582. Initial value. (Contributed by Jim Kingdon, 15-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    ⇒   ⊢ (𝜑 → (𝐻‘0) = ∅)
Theorem	ennnfonelemp1 12563*	Lemma for ennnfone 12582. Value of 𝐻 at a successor. (Contributed by Jim Kingdon, 23-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝐻‘(𝑃 + 1)) = if((𝐹‘(◡𝑁‘𝑃)) ∈ (𝐹 “ (◡𝑁‘𝑃)), (𝐻‘𝑃), ((𝐻‘𝑃) ∪ {⟨dom (𝐻‘𝑃), (𝐹‘(◡𝑁‘𝑃))⟩})))
Theorem	ennnfonelem1 12564*	Lemma for ennnfone 12582. Second value. (Contributed by Jim Kingdon, 19-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    ⇒   ⊢ (𝜑 → (𝐻‘1) = {⟨∅, (𝐹‘∅)⟩})
Theorem	ennnfonelemom 12565*	Lemma for ennnfone 12582. 𝐻 yields finite sequences. (Contributed by Jim Kingdon, 19-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    ⇒   ⊢ (𝜑 → dom (𝐻‘𝑃) ∈ ω)
Theorem	ennnfonelemhdmp1 12566*	Lemma for ennnfone 12582. Domain at a successor where we need to add an element to the sequence. (Contributed by Jim Kingdon, 23-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    &   ⊢ (𝜑 → ¬ (𝐹‘(◡𝑁‘𝑃)) ∈ (𝐹 “ (◡𝑁‘𝑃)))    ⇒   ⊢ (𝜑 → dom (𝐻‘(𝑃 + 1)) = suc dom (𝐻‘𝑃))
Theorem	ennnfonelemss 12567*	Lemma for ennnfone 12582. We only add elements to 𝐻 as the index increases. (Contributed by Jim Kingdon, 15-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝐻‘𝑃) ⊆ (𝐻‘(𝑃 + 1)))
Theorem	ennnfoneleminc 12568*	Lemma for ennnfone 12582. We only add elements to 𝐻 as the index increases. (Contributed by Jim Kingdon, 21-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑄 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑃 ≤ 𝑄)    ⇒   ⊢ (𝜑 → (𝐻‘𝑃) ⊆ (𝐻‘𝑄))
Theorem	ennnfonelemkh 12569*	Lemma for ennnfone 12582. Because we add zero or one entries for each new index, the length of each sequence is no greater than its index. (Contributed by Jim Kingdon, 19-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    ⇒   ⊢ (𝜑 → dom (𝐻‘𝑃) ⊆ (◡𝑁‘𝑃))
Theorem	ennnfonelemhf1o 12570*	Lemma for ennnfone 12582. Each of the functions in 𝐻 is one to one and onto an image of 𝐹. (Contributed by Jim Kingdon, 17-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝐻‘𝑃):dom (𝐻‘𝑃)–1-1-onto→(𝐹 “ (◡𝑁‘𝑃)))
Theorem	ennnfonelemex 12571*	Lemma for ennnfone 12582. Extending the sequence (𝐻‘𝑃) to include an additional element. (Contributed by Jim Kingdon, 19-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    ⇒   ⊢ (𝜑 → ∃𝑖 ∈ ℕ0 dom (𝐻‘𝑃) ∈ dom (𝐻‘𝑖))
Theorem	ennnfonelemhom 12572*	Lemma for ennnfone 12582. The sequences in 𝐻 increase in length without bound if you go out far enough. (Contributed by Jim Kingdon, 19-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑀 ∈ ω)    ⇒   ⊢ (𝜑 → ∃𝑖 ∈ ℕ0 𝑀 ∈ dom (𝐻‘𝑖))
Theorem	ennnfonelemrnh 12573*	Lemma for ennnfone 12582. A consequence of ennnfonelemss 12567. (Contributed by Jim Kingdon, 16-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑋 ∈ ran 𝐻)    &   ⊢ (𝜑 → 𝑌 ∈ ran 𝐻)    ⇒   ⊢ (𝜑 → (𝑋 ⊆ 𝑌 ∨ 𝑌 ⊆ 𝑋))
Theorem	ennnfonelemfun 12574*	Lemma for ennnfone 12582. 𝐿 is a function. (Contributed by Jim Kingdon, 16-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ 𝐿 = ∪ 𝑖 ∈ ℕ0 (𝐻‘𝑖)    ⇒   ⊢ (𝜑 → Fun 𝐿)
Theorem	ennnfonelemf1 12575*	Lemma for ennnfone 12582. 𝐿 is one-to-one. (Contributed by Jim Kingdon, 16-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ 𝐿 = ∪ 𝑖 ∈ ℕ0 (𝐻‘𝑖)    ⇒   ⊢ (𝜑 → 𝐿:dom 𝐿–1-1→𝐴)
Theorem	ennnfonelemrn 12576*	Lemma for ennnfone 12582. 𝐿 is onto 𝐴. (Contributed by Jim Kingdon, 16-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ 𝐿 = ∪ 𝑖 ∈ ℕ0 (𝐻‘𝑖)    ⇒   ⊢ (𝜑 → ran 𝐿 = 𝐴)
Theorem	ennnfonelemdm 12577*	Lemma for ennnfone 12582. The function 𝐿 is defined everywhere. (Contributed by Jim Kingdon, 16-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ 𝐿 = ∪ 𝑖 ∈ ℕ0 (𝐻‘𝑖)    ⇒   ⊢ (𝜑 → dom 𝐿 = ω)
Theorem	ennnfonelemen 12578*	Lemma for ennnfone 12582. The result. (Contributed by Jim Kingdon, 16-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ 𝐿 = ∪ 𝑖 ∈ ℕ0 (𝐻‘𝑖)    ⇒   ⊢ (𝜑 → 𝐴 ≈ ℕ)
Theorem	ennnfonelemnn0 12579*	Lemma for ennnfone 12582. A version of ennnfonelemen 12578 expressed in terms of ℕ0 instead of ω. (Contributed by Jim Kingdon, 27-Oct-2022.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ℕ0–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ℕ0 ∃𝑘 ∈ ℕ0 ∀𝑗 ∈ (0...𝑛)(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    ⇒   ⊢ (𝜑 → 𝐴 ≈ ℕ)
Theorem	ennnfonelemr 12580*	Lemma for ennnfone 12582. The interesting direction, expressed in deduction form. (Contributed by Jim Kingdon, 27-Oct-2022.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ℕ0–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ℕ0 ∃𝑘 ∈ ℕ0 ∀𝑗 ∈ (0...𝑛)(𝐹‘𝑘) ≠ (𝐹‘𝑗))    ⇒   ⊢ (𝜑 → 𝐴 ≈ ℕ)
Theorem	ennnfonelemim 12581*	Lemma for ennnfone 12582. The trivial direction. (Contributed by Jim Kingdon, 27-Oct-2022.)
⊢ (𝐴 ≈ ℕ → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦 ∧ ∃𝑓(𝑓:ℕ0–onto→𝐴 ∧ ∀𝑛 ∈ ℕ0 ∃𝑘 ∈ ℕ0 ∀𝑗 ∈ (0...𝑛)(𝑓‘𝑘) ≠ (𝑓‘𝑗))))
Theorem	ennnfone 12582*	A condition for a set being countably infinite. Corollary 8.1.13 of [AczelRathjen], p. 73. Roughly speaking, the condition says that 𝐴 is countable (that's the 𝑓:ℕ0–onto→𝐴 part, as seen in theorems like ctm 7168), infinite (that's the part about being able to find an element of 𝐴 distinct from any mapping of a natural number via 𝑓), and has decidable equality. (Contributed by Jim Kingdon, 27-Oct-2022.)
⊢ (𝐴 ≈ ℕ ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦 ∧ ∃𝑓(𝑓:ℕ0–onto→𝐴 ∧ ∀𝑛 ∈ ℕ0 ∃𝑘 ∈ ℕ0 ∀𝑗 ∈ (0...𝑛)(𝑓‘𝑘) ≠ (𝑓‘𝑗))))
Theorem	exmidunben 12583*	If any unbounded set of positive integers is equinumerous to ℕ, then the Limited Principle of Omniscience (LPO) implies excluded middle. (Contributed by Jim Kingdon, 29-Jul-2023.)
⊢ ((∀𝑥((𝑥 ⊆ ℕ ∧ ∀𝑚 ∈ ℕ ∃𝑛 ∈ 𝑥 𝑚 < 𝑛) → 𝑥 ≈ ℕ) ∧ ω ∈ Omni) → EXMID)
Theorem	ctinfomlemom 12584*	Lemma for ctinfom 12585. Converting between ω and ℕ0. (Contributed by Jim Kingdon, 10-Aug-2023.)
⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐺 = (𝐹 ∘ ◡𝑁)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ¬ (𝐹‘𝑘) ∈ (𝐹 “ 𝑛))    ⇒   ⊢ (𝜑 → (𝐺:ℕ0–onto→𝐴 ∧ ∀𝑚 ∈ ℕ0 ∃𝑗 ∈ ℕ0 ∀𝑖 ∈ (0...𝑚)(𝐺‘𝑗) ≠ (𝐺‘𝑖)))
Theorem	ctinfom 12585*	A condition for a set being countably infinite. Restates ennnfone 12582 in terms of ω and function image. Like ennnfone 12582 the condition can be summarized as 𝐴 being countable, infinite, and having decidable equality. (Contributed by Jim Kingdon, 7-Aug-2023.)
⊢ (𝐴 ≈ ℕ ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦 ∧ ∃𝑓(𝑓:ω–onto→𝐴 ∧ ∀𝑛 ∈ ω ∃𝑘 ∈ ω ¬ (𝑓‘𝑘) ∈ (𝑓 “ 𝑛))))
Theorem	inffinp1 12586*	An infinite set contains an element not contained in a given finite subset. (Contributed by Jim Kingdon, 7-Aug-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → ω ≼ 𝐴)    &   ⊢ (𝜑 → 𝐵 ⊆ 𝐴)    &   ⊢ (𝜑 → 𝐵 ∈ Fin)    ⇒   ⊢ (𝜑 → ∃𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵)
Theorem	ctinf 12587*	A set is countably infinite if and only if it has decidable equality, is countable, and is infinite. (Contributed by Jim Kingdon, 7-Aug-2023.)
⊢ (𝐴 ≈ ℕ ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦 ∧ ∃𝑓 𝑓:ω–onto→𝐴 ∧ ω ≼ 𝐴))
Theorem	qnnen 12588	The rational numbers are countably infinite. Corollary 8.1.23 of [AczelRathjen], p. 75. This is Metamath 100 proof #3. (Contributed by Jim Kingdon, 11-Aug-2023.)
⊢ ℚ ≈ ℕ
Theorem	enctlem 12589*	Lemma for enct 12590. One direction of the biconditional. (Contributed by Jim Kingdon, 23-Dec-2023.)
⊢ (𝐴 ≈ 𝐵 → (∃𝑓 𝑓:ω–onto→(𝐴 ⊔ 1o) → ∃𝑔 𝑔:ω–onto→(𝐵 ⊔ 1o)))
Theorem	enct 12590*	Countability is invariant relative to equinumerosity. (Contributed by Jim Kingdon, 23-Dec-2023.)
⊢ (𝐴 ≈ 𝐵 → (∃𝑓 𝑓:ω–onto→(𝐴 ⊔ 1o) ↔ ∃𝑔 𝑔:ω–onto→(𝐵 ⊔ 1o)))
Theorem	ctiunctlemu1st 12591*	Lemma for ctiunct 12597. (Contributed by Jim Kingdon, 28-Oct-2023.)
⊢ (𝜑 → 𝑆 ⊆ ω)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω DECID 𝑛 ∈ 𝑆)    &   ⊢ (𝜑 → 𝐹:𝑆–onto→𝐴)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑇 ⊆ ω)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → ∀𝑛 ∈ ω DECID 𝑛 ∈ 𝑇)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐺:𝑇–onto→𝐵)    &   ⊢ (𝜑 → 𝐽:ω–1-1-onto→(ω × ω))    &   ⊢ 𝑈 = {𝑧 ∈ ω ∣ ((1st ‘(𝐽‘𝑧)) ∈ 𝑆 ∧ (2nd ‘(𝐽‘𝑧)) ∈ ⦋(𝐹‘(1st ‘(𝐽‘𝑧))) / 𝑥⦌𝑇)}    &   ⊢ (𝜑 → 𝑁 ∈ 𝑈)    ⇒   ⊢ (𝜑 → (1st ‘(𝐽‘𝑁)) ∈ 𝑆)
Theorem	ctiunctlemu2nd 12592*	Lemma for ctiunct 12597. (Contributed by Jim Kingdon, 28-Oct-2023.)
⊢ (𝜑 → 𝑆 ⊆ ω)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω DECID 𝑛 ∈ 𝑆)    &   ⊢ (𝜑 → 𝐹:𝑆–onto→𝐴)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑇 ⊆ ω)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → ∀𝑛 ∈ ω DECID 𝑛 ∈ 𝑇)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐺:𝑇–onto→𝐵)    &   ⊢ (𝜑 → 𝐽:ω–1-1-onto→(ω × ω))    &   ⊢ 𝑈 = {𝑧 ∈ ω ∣ ((1st ‘(𝐽‘𝑧)) ∈ 𝑆 ∧ (2nd ‘(𝐽‘𝑧)) ∈ ⦋(𝐹‘(1st ‘(𝐽‘𝑧))) / 𝑥⦌𝑇)}    &   ⊢ (𝜑 → 𝑁 ∈ 𝑈)    ⇒   ⊢ (𝜑 → (2nd ‘(𝐽‘𝑁)) ∈ ⦋(𝐹‘(1st ‘(𝐽‘𝑁))) / 𝑥⦌𝑇)
Theorem	ctiunctlemuom 12593	Lemma for ctiunct 12597. (Contributed by Jim Kingdon, 28-Oct-2023.)
⊢ (𝜑 → 𝑆 ⊆ ω)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω DECID 𝑛 ∈ 𝑆)    &   ⊢ (𝜑 → 𝐹:𝑆–onto→𝐴)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑇 ⊆ ω)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → ∀𝑛 ∈ ω DECID 𝑛 ∈ 𝑇)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐺:𝑇–onto→𝐵)    &   ⊢ (𝜑 → 𝐽:ω–1-1-onto→(ω × ω))    &   ⊢ 𝑈 = {𝑧 ∈ ω ∣ ((1st ‘(𝐽‘𝑧)) ∈ 𝑆 ∧ (2nd ‘(𝐽‘𝑧)) ∈ ⦋(𝐹‘(1st ‘(𝐽‘𝑧))) / 𝑥⦌𝑇)}    ⇒   ⊢ (𝜑 → 𝑈 ⊆ ω)
Theorem	ctiunctlemudc 12594*	Lemma for ctiunct 12597. (Contributed by Jim Kingdon, 28-Oct-2023.)
⊢ (𝜑 → 𝑆 ⊆ ω)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω DECID 𝑛 ∈ 𝑆)    &   ⊢ (𝜑 → 𝐹:𝑆–onto→𝐴)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑇 ⊆ ω)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → ∀𝑛 ∈ ω DECID 𝑛 ∈ 𝑇)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐺:𝑇–onto→𝐵)    &   ⊢ (𝜑 → 𝐽:ω–1-1-onto→(ω × ω))    &   ⊢ 𝑈 = {𝑧 ∈ ω ∣ ((1st ‘(𝐽‘𝑧)) ∈ 𝑆 ∧ (2nd ‘(𝐽‘𝑧)) ∈ ⦋(𝐹‘(1st ‘(𝐽‘𝑧))) / 𝑥⦌𝑇)}    ⇒   ⊢ (𝜑 → ∀𝑛 ∈ ω DECID 𝑛 ∈ 𝑈)
Theorem	ctiunctlemf 12595*	Lemma for ctiunct 12597. (Contributed by Jim Kingdon, 28-Oct-2023.)
⊢ (𝜑 → 𝑆 ⊆ ω)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω DECID 𝑛 ∈ 𝑆)    &   ⊢ (𝜑 → 𝐹:𝑆–onto→𝐴)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑇 ⊆ ω)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → ∀𝑛 ∈ ω DECID 𝑛 ∈ 𝑇)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐺:𝑇–onto→𝐵)    &   ⊢ (𝜑 → 𝐽:ω–1-1-onto→(ω × ω))    &   ⊢ 𝑈 = {𝑧 ∈ ω ∣ ((1st ‘(𝐽‘𝑧)) ∈ 𝑆 ∧ (2nd ‘(𝐽‘𝑧)) ∈ ⦋(𝐹‘(1st ‘(𝐽‘𝑧))) / 𝑥⦌𝑇)}    &   ⊢ 𝐻 = (𝑛 ∈ 𝑈 ↦ (⦋(𝐹‘(1st ‘(𝐽‘𝑛))) / 𝑥⦌𝐺‘(2nd ‘(𝐽‘𝑛))))    ⇒   ⊢ (𝜑 → 𝐻:𝑈⟶∪ 𝑥 ∈ 𝐴 𝐵)
Theorem	ctiunctlemfo 12596*	Lemma for ctiunct 12597. (Contributed by Jim Kingdon, 28-Oct-2023.)
⊢ (𝜑 → 𝑆 ⊆ ω)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω DECID 𝑛 ∈ 𝑆)    &   ⊢ (𝜑 → 𝐹:𝑆–onto→𝐴)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑇 ⊆ ω)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → ∀𝑛 ∈ ω DECID 𝑛 ∈ 𝑇)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐺:𝑇–onto→𝐵)    &   ⊢ (𝜑 → 𝐽:ω–1-1-onto→(ω × ω))    &   ⊢ 𝑈 = {𝑧 ∈ ω ∣ ((1st ‘(𝐽‘𝑧)) ∈ 𝑆 ∧ (2nd ‘(𝐽‘𝑧)) ∈ ⦋(𝐹‘(1st ‘(𝐽‘𝑧))) / 𝑥⦌𝑇)}    &   ⊢ 𝐻 = (𝑛 ∈ 𝑈 ↦ (⦋(𝐹‘(1st ‘(𝐽‘𝑛))) / 𝑥⦌𝐺‘(2nd ‘(𝐽‘𝑛))))    &   ⊢ Ⅎ𝑥𝐻    &   ⊢ Ⅎ𝑥𝑈    ⇒   ⊢ (𝜑 → 𝐻:𝑈–onto→∪ 𝑥 ∈ 𝐴 𝐵)
Theorem	ctiunct 12597*	A sequence of enumerations gives an enumeration of the union. We refer to "sequence of enumerations" rather than "countably many countable sets" because the hypothesis provides more than countability for each 𝐵(𝑥): it refers to 𝐵(𝑥) together with the 𝐺(𝑥) which enumerates it. Theorem 8.1.19 of [AczelRathjen], p. 74. For "countably many countable sets" the key hypothesis would be (𝜑 ∧ 𝑥 ∈ 𝐴) → ∃𝑔𝑔:ω–onto→(𝐵 ⊔ 1o). This is almost omiunct 12601 (which uses countable choice) although that is for a countably infinite collection not any countable collection. Compare with the case of two sets instead of countably many, as seen at unct 12599, which says that the union of two countable sets is countable . The proof proceeds by mapping a natural number to a pair of natural numbers (by xpomen 12552) and using the first number to map to an element 𝑥 of 𝐴 and the second number to map to an element of B(x) . In this way we are able to map to every element of ∪ 𝑥 ∈ 𝐴𝐵. Although it would be possible to work directly with countability expressed as 𝐹:ω–onto→(𝐴 ⊔ 1o), we instead use functions from subsets of the natural numbers via ctssdccl 7170 and ctssdc 7172. (Contributed by Jim Kingdon, 31-Oct-2023.)
⊢ (𝜑 → 𝐹:ω–onto→(𝐴 ⊔ 1o))    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐺:ω–onto→(𝐵 ⊔ 1o))    ⇒   ⊢ (𝜑 → ∃ℎ ℎ:ω–onto→(∪ 𝑥 ∈ 𝐴 𝐵 ⊔ 1o))
Theorem	ctiunctal 12598*	Variation of ctiunct 12597 which allows 𝑥 to be present in 𝜑. (Contributed by Jim Kingdon, 5-May-2024.)
⊢ (𝜑 → 𝐹:ω–onto→(𝐴 ⊔ 1o))    &   ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 𝐺:ω–onto→(𝐵 ⊔ 1o))    ⇒   ⊢ (𝜑 → ∃ℎ ℎ:ω–onto→(∪ 𝑥 ∈ 𝐴 𝐵 ⊔ 1o))
Theorem	unct 12599*	The union of two countable sets is countable. Corollary 8.1.20 of [AczelRathjen], p. 75. (Contributed by Jim Kingdon, 1-Nov-2023.)
⊢ ((∃𝑓 𝑓:ω–onto→(𝐴 ⊔ 1o) ∧ ∃𝑔 𝑔:ω–onto→(𝐵 ⊔ 1o)) → ∃ℎ ℎ:ω–onto→((𝐴 ∪ 𝐵) ⊔ 1o))
Theorem	omctfn 12600*	Using countable choice to find a sequence of enumerations for a collection of countable sets. Lemma 8.1.27 of [AczelRathjen], p. 77. (Contributed by Jim Kingdon, 19-Apr-2024.)
⊢ (𝜑 → CCHOICE)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ω) → ∃𝑔 𝑔:ω–onto→(𝐵 ⊔ 1o))    ⇒   ⊢ (𝜑 → ∃𝑓(𝑓 Fn ω ∧ ∀𝑥 ∈ ω (𝑓‘𝑥):ω–onto→(𝐵 ⊔ 1o)))