P. 130 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 12901-13000   *Has distinct variable group(s)

Type	Label	Description
Statement
Definition	df-gz 12901	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 12902	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 12903	A gaussian integer is a complex number. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → 𝐴 ∈ ℂ)
Theorem	zgz 12904	An integer is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ → 𝐴 ∈ ℤ[i])
Theorem	igz 12905	i is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ i ∈ ℤ[i]
Theorem	gznegcl 12906	The gaussian integers are closed under negation. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → -𝐴 ∈ ℤ[i])
Theorem	gzcjcl 12907	The gaussian integers are closed under conjugation. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → (∗‘𝐴) ∈ ℤ[i])
Theorem	gzaddcl 12908	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 12909	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 12910	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 12911	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 12912	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 12913	Lemma for 4sq 12941. (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 12914	Lemma for 4sq 12941. (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 12915	Lemma for 4sq 12941. (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 12916	Lemma for 4sq 12941. (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 12917	Lemma for 4sq 12941. (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 12918	Lemma for 4sq 12941. (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 12919*	Lemma for 4sq 12941. 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 12920*	Lemma for 4sq 12941. 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 12921*	Lemma for 4sq 12941. 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 12922*	Lemma for 4sqlem4 12923. (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 12923*	Lemma for 4sq 12941. 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 12924*	Lemma for mul4sq 12925: 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 12925*	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 12924. (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 12926*	Lemma for 4sq 12941. 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 12927*	Lemma for 4sq 12941. 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 12928*	Lemma for 4sq 12941. 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 12929*	Exercise which may help in understanding the proof of 4sqlemsdc 12931. (Contributed by Jim Kingdon, 25-May-2025.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ 𝑛 = (𝑥↑2)}    ⇒   ⊢ (𝐴 ∈ ℕ0 → DECID 𝐴 ∈ 𝑆)
Theorem	4sqexercise2 12930*	Exercise which may help in understanding the proof of 4sqlemsdc 12931. (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 12931*	Lemma for 4sq 12941. 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 12929 and 4sqexercise2 12930 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 12932*	Lemma for 4sq 12941. 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 12928 and fin0 7055. (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 12933*	Lemma for 4sq 12941. 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 12934*	Lemma for 4sq 12941. (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 12935*	Lemma for 4sq 12941. (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 12936*	Lemma for 4sq 12941. (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 12937*	Lemma for 4sq 12941. (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 12938*	Lemma for 4sq 12941. (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 12939*	Lemma for 4sq 12941. 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 12940*	Lemma for 4sq 12941. 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 12939. 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 12925 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 12941*	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))))
5.2.13  Decimal arithmetic (cont.)
Theorem	dec2dvds 12942	Divisibility by two is obvious in base 10. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ (𝐵 · 2) = 𝐶    &   ⊢ 𝐷 = (𝐶 + 1)    ⇒   ⊢ ¬ 2 ∥ ;𝐴𝐷
Theorem	dec5dvds 12943	Divisibility by five is obvious in base 10. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ    &   ⊢ 𝐵 < 5    ⇒   ⊢ ¬ 5 ∥ ;𝐴𝐵
Theorem	dec5dvds2 12944	Divisibility by five is obvious in base 10. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ    &   ⊢ 𝐵 < 5    &   ⊢ (5 + 𝐵) = 𝐶    ⇒   ⊢ ¬ 5 ∥ ;𝐴𝐶
Theorem	dec5nprm 12945	A decimal number greater than 10 and ending with five is not a prime number. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ    ⇒   ⊢ ¬ ;𝐴5 ∈ ℙ
Theorem	dec2nprm 12946	A decimal number greater than 10 and ending with an even digit is not a prime number. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ (𝐵 · 2) = 𝐶    ⇒   ⊢ ¬ ;𝐴𝐶 ∈ ℙ
Theorem	modxai 12947	Add exponents in a power mod calculation. (Contributed by Mario Carneiro, 21-Feb-2014.) (Revised by Mario Carneiro, 5-Feb-2015.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ 𝐶 ∈ ℕ0    &   ⊢ 𝐿 ∈ ℕ0    &   ⊢ ((𝐴↑𝐵) mod 𝑁) = (𝐾 mod 𝑁)    &   ⊢ ((𝐴↑𝐶) mod 𝑁) = (𝐿 mod 𝑁)    &   ⊢ (𝐵 + 𝐶) = 𝐸    &   ⊢ ((𝐷 · 𝑁) + 𝑀) = (𝐾 · 𝐿)    ⇒   ⊢ ((𝐴↑𝐸) mod 𝑁) = (𝑀 mod 𝑁)
Theorem	mod2xi 12948	Double exponents in a power mod calculation. (Contributed by Mario Carneiro, 21-Feb-2014.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ ((𝐴↑𝐵) mod 𝑁) = (𝐾 mod 𝑁)    &   ⊢ (2 · 𝐵) = 𝐸    &   ⊢ ((𝐷 · 𝑁) + 𝑀) = (𝐾 · 𝐾)    ⇒   ⊢ ((𝐴↑𝐸) mod 𝑁) = (𝑀 mod 𝑁)
Theorem	modxp1i 12949	Add one to an exponent in a power mod calculation. (Contributed by Mario Carneiro, 21-Feb-2014.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ ((𝐴↑𝐵) mod 𝑁) = (𝐾 mod 𝑁)    &   ⊢ (𝐵 + 1) = 𝐸    &   ⊢ ((𝐷 · 𝑁) + 𝑀) = (𝐾 · 𝐴)    ⇒   ⊢ ((𝐴↑𝐸) mod 𝑁) = (𝑀 mod 𝑁)
Theorem	modsubi 12950	Subtract from within a mod calculation. (Contributed by Mario Carneiro, 18-Feb-2014.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (𝐴 mod 𝑁) = (𝐾 mod 𝑁)    &   ⊢ (𝑀 + 𝐵) = 𝐾    ⇒   ⊢ ((𝐴 − 𝐵) mod 𝑁) = (𝑀 mod 𝑁)
Theorem	gcdi 12951	Calculate a GCD via Euclid's algorithm. (Contributed by Mario Carneiro, 19-Feb-2014.)
⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑅 ∈ ℕ0    &   ⊢ 𝑁 ∈ ℕ0    &   ⊢ (𝑁 gcd 𝑅) = 𝐺    &   ⊢ ((𝐾 · 𝑁) + 𝑅) = 𝑀    ⇒   ⊢ (𝑀 gcd 𝑁) = 𝐺
Theorem	gcdmodi 12952	Calculate a GCD via Euclid's algorithm. Theorem 5.6 in [ApostolNT] p. 109. (Contributed by Mario Carneiro, 19-Feb-2014.)
⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑅 ∈ ℕ0    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ (𝐾 mod 𝑁) = (𝑅 mod 𝑁)    &   ⊢ (𝑁 gcd 𝑅) = 𝐺    ⇒   ⊢ (𝐾 gcd 𝑁) = 𝐺
Theorem	numexp0 12953	Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ (𝐴↑0) = 1
Theorem	numexp1 12954	Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ (𝐴↑1) = 𝐴
Theorem	numexpp1 12955	Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (𝑀 + 1) = 𝑁    &   ⊢ ((𝐴↑𝑀) · 𝐴) = 𝐶    ⇒   ⊢ (𝐴↑𝑁) = 𝐶
Theorem	numexp2x 12956	Double an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (2 · 𝑀) = 𝑁    &   ⊢ (𝐴↑𝑀) = 𝐷    &   ⊢ (𝐷 · 𝐷) = 𝐶    ⇒   ⊢ (𝐴↑𝑁) = 𝐶
Theorem	decsplit0b 12957	Split a decimal number into two parts. Base case: 𝑁 = 0. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 8-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ ((𝐴 · (;10↑0)) + 𝐵) = (𝐴 + 𝐵)
Theorem	decsplit0 12958	Split a decimal number into two parts. Base case: 𝑁 = 0. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 8-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ ((𝐴 · (;10↑0)) + 0) = 𝐴
Theorem	decsplit1 12959	Split a decimal number into two parts. Base case: 𝑁 = 1. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 8-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ ((𝐴 · (;10↑1)) + 𝐵) = ;𝐴𝐵
Theorem	decsplit 12960	Split a decimal number into two parts. Inductive step. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 8-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (𝑀 + 1) = 𝑁    &   ⊢ ((𝐴 · (;10↑𝑀)) + 𝐵) = 𝐶    ⇒   ⊢ ((𝐴 · (;10↑𝑁)) + ;𝐵𝐷) = ;𝐶𝐷
Theorem	karatsuba 12961	The Karatsuba multiplication algorithm. If 𝑋 and 𝑌 are decomposed into two groups of digits of length 𝑀 (only the lower group is known to be this size but the algorithm is most efficient when the partition is chosen near the middle of the digit string), then 𝑋𝑌 can be written in three groups of digits, where each group needs only one multiplication. Thus, we can halve both inputs with only three multiplications on the smaller operands, yielding an asymptotic improvement of n^(log2 3) instead of n^2 for the "naive" algorithm decmul1c 9650. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 9-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐶 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℕ0    &   ⊢ 𝑆 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (𝐴 · 𝐶) = 𝑅    &   ⊢ (𝐵 · 𝐷) = 𝑇    &   ⊢ ((𝐴 + 𝐵) · (𝐶 + 𝐷)) = ((𝑅 + 𝑆) + 𝑇)    &   ⊢ ((𝐴 · (;10↑𝑀)) + 𝐵) = 𝑋    &   ⊢ ((𝐶 · (;10↑𝑀)) + 𝐷) = 𝑌    &   ⊢ ((𝑅 · (;10↑𝑀)) + 𝑆) = 𝑊    &   ⊢ ((𝑊 · (;10↑𝑀)) + 𝑇) = 𝑍    ⇒   ⊢ (𝑋 · 𝑌) = 𝑍
Theorem	2exp4 12962	Two to the fourth power is 16. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (2↑4) = ;16
Theorem	2exp5 12963	Two to the fifth power is 32. (Contributed by AV, 16-Aug-2021.)
⊢ (2↑5) = ;32
Theorem	2exp6 12964	Two to the sixth power is 64. (Contributed by Mario Carneiro, 20-Apr-2015.) (Proof shortened by OpenAI, 25-Mar-2020.)
⊢ (2↑6) = ;64
Theorem	2exp7 12965	Two to the seventh power is 128. (Contributed by AV, 16-Aug-2021.)
⊢ (2↑7) = ;;128
Theorem	2exp8 12966	Two to the eighth power is 256. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (2↑8) = ;;256
Theorem	2exp11 12967	Two to the eleventh power is 2048. (Contributed by AV, 16-Aug-2021.)
⊢ (2↑;11) = ;;;2048
Theorem	2exp16 12968	Two to the sixteenth power is 65536. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (2↑;16) = ;;;;65536
Theorem	3exp3 12969	Three to the third power is 27. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (3↑3) = ;27
Theorem	2expltfac 12970	The factorial grows faster than two to the power 𝑁. (Contributed by Mario Carneiro, 15-Sep-2016.)
⊢ (𝑁 ∈ (ℤ≥‘4) → (2↑𝑁) < (!‘𝑁))
5.3  Cardinality of real and complex number subsets
5.3.1  Countability of integers and rationals
Theorem	oddennn 12971	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 12972	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 12973	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 12974	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 12975	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 12976	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 12977	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 12978*	Lemma for ennnfone 13004. A direct consequence of fidcenumlemrk 7129. (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 12979*	Lemma for ennnfone 13004. (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 12980*	Lemma for ennnfone 13004. 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 12981*	Lemma for ennnfone 13004. 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 12982*	Lemma for ennnfone 13004. 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 12983*	Lemma for ennnfone 13004. (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 12984*	Lemma for ennnfone 13004. 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 12985*	Lemma for ennnfone 13004. 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 12986*	Lemma for ennnfone 13004. 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 12987*	Lemma for ennnfone 13004. 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 12988*	Lemma for ennnfone 13004. 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 12989*	Lemma for ennnfone 13004. 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 12990*	Lemma for ennnfone 13004. 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 12991*	Lemma for ennnfone 13004. 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 12992*	Lemma for ennnfone 13004. 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 12993*	Lemma for ennnfone 13004. 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 12994*	Lemma for ennnfone 13004. 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 12995*	Lemma for ennnfone 13004. A consequence of ennnfonelemss 12989. (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 12996*	Lemma for ennnfone 13004. 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 12997*	Lemma for ennnfone 13004. 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 12998*	Lemma for ennnfone 13004. 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 12999*	Lemma for ennnfone 13004. 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 13000*	Lemma for ennnfone 13004. 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)