P. 477 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 47601-47700   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	sigarexp 47601*	Expand the signed area formula by linearity. (Contributed by Saveliy Skresanov, 20-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ((𝐴 − 𝐶)𝐺(𝐵 − 𝐶)) = (((𝐴𝐺𝐵) − (𝐴𝐺𝐶)) − (𝐶𝐺𝐵)))
Theorem	sigarperm 47602*	Signed area (𝐴 − 𝐶)𝐺(𝐵 − 𝐶) acts as a double area of a triangle 𝐴𝐵𝐶. Here we prove that cyclically permuting the vertices doesn't change the area. (Contributed by Saveliy Skresanov, 20-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ((𝐴 − 𝐶)𝐺(𝐵 − 𝐶)) = ((𝐵 − 𝐴)𝐺(𝐶 − 𝐴)))
Theorem	sigardiv 47603*	If signed area between vectors 𝐵 − 𝐴 and 𝐶 − 𝐴 is zero, then those vectors lie on the same line. (Contributed by Saveliy Skresanov, 22-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    &   ⊢ (𝜑 → (𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ))    &   ⊢ (𝜑 → ¬ 𝐶 = 𝐴)    &   ⊢ (𝜑 → ((𝐵 − 𝐴)𝐺(𝐶 − 𝐴)) = 0)    ⇒   ⊢ (𝜑 → ((𝐵 − 𝐴) / (𝐶 − 𝐴)) ∈ ℝ)
Theorem	sigarimcd 47604*	Signed area takes value in complex numbers. Deduction version. (Contributed by Saveliy Skresanov, 23-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    &   ⊢ (𝜑 → (𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ))    ⇒   ⊢ (𝜑 → (𝐴𝐺𝐵) ∈ ℂ)
Theorem	sigariz 47605*	If signed area is zero, the signed area with swapped arguments is also zero. Deduction version. (Contributed by Saveliy Skresanov, 23-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    &   ⊢ (𝜑 → (𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ))    &   ⊢ (𝜑 → (𝐴𝐺𝐵) = 0)    ⇒   ⊢ (𝜑 → (𝐵𝐺𝐴) = 0)
Theorem	sigarcol 47606*	Given three points 𝐴, 𝐵 and 𝐶 such that ¬ 𝐴 = 𝐵, the point 𝐶 lies on the line going through 𝐴 and 𝐵 iff the corresponding signed area is zero. That justifies the usage of signed area as a collinearity indicator. (Contributed by Saveliy Skresanov, 22-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    &   ⊢ (𝜑 → (𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ))    &   ⊢ (𝜑 → ¬ 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → (((𝐴 − 𝐶)𝐺(𝐵 − 𝐶)) = 0 ↔ ∃𝑡 ∈ ℝ 𝐶 = (𝐵 + (𝑡 · (𝐴 − 𝐵)))))
Theorem	sharhght 47607*	Let 𝐴𝐵𝐶 be a triangle, and let 𝐷 lie on the line 𝐴𝐵. Then (doubled) areas of triangles 𝐴𝐷𝐶 and 𝐶𝐷𝐵 relate as lengths of corresponding bases 𝐴𝐷 and 𝐷𝐵. (Contributed by Saveliy Skresanov, 23-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    &   ⊢ (𝜑 → (𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ))    &   ⊢ (𝜑 → (𝐷 ∈ ℂ ∧ ((𝐴 − 𝐷)𝐺(𝐵 − 𝐷)) = 0))    ⇒   ⊢ (𝜑 → (((𝐶 − 𝐴)𝐺(𝐷 − 𝐴)) · (𝐵 − 𝐷)) = (((𝐶 − 𝐵)𝐺(𝐷 − 𝐵)) · (𝐴 − 𝐷)))
Theorem	sigaradd 47608*	Subtracting (double) area of 𝐴𝐷𝐶 from 𝐴𝐵𝐶 yields the (double) area of 𝐷𝐵𝐶. (Contributed by Saveliy Skresanov, 23-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    &   ⊢ (𝜑 → (𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ))    &   ⊢ (𝜑 → (𝐷 ∈ ℂ ∧ ((𝐴 − 𝐷)𝐺(𝐵 − 𝐷)) = 0))    ⇒   ⊢ (𝜑 → (((𝐵 − 𝐶)𝐺(𝐴 − 𝐶)) − ((𝐷 − 𝐶)𝐺(𝐴 − 𝐶))) = ((𝐵 − 𝐶)𝐺(𝐷 − 𝐶)))
Theorem	cevathlem1 47609	Ceva's theorem first lemma. Multiplies three identities and divides by the common factors. (Contributed by Saveliy Skresanov, 24-Sep-2017.)
⊢ (𝜑 → (𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ))    &   ⊢ (𝜑 → (𝐷 ∈ ℂ ∧ 𝐸 ∈ ℂ ∧ 𝐹 ∈ ℂ))    &   ⊢ (𝜑 → (𝐺 ∈ ℂ ∧ 𝐻 ∈ ℂ ∧ 𝐾 ∈ ℂ))    &   ⊢ (𝜑 → (𝐴 ≠ 0 ∧ 𝐸 ≠ 0 ∧ 𝐶 ≠ 0))    &   ⊢ (𝜑 → ((𝐴 · 𝐵) = (𝐶 · 𝐷) ∧ (𝐸 · 𝐹) = (𝐴 · 𝐺) ∧ (𝐶 · 𝐻) = (𝐸 · 𝐾)))    ⇒   ⊢ (𝜑 → ((𝐵 · 𝐹) · 𝐻) = ((𝐷 · 𝐺) · 𝐾))
Theorem	cevathlem2 47610*	Ceva's theorem second lemma. Relate (doubled) areas of triangles 𝐶𝐴𝑂 and 𝐴𝐵𝑂 with of segments 𝐵𝐷 and 𝐷𝐶. (Contributed by Saveliy Skresanov, 24-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    &   ⊢ (𝜑 → (𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ))    &   ⊢ (𝜑 → (𝐹 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐸 ∈ ℂ))    &   ⊢ (𝜑 → 𝑂 ∈ ℂ)    &   ⊢ (𝜑 → (((𝐴 − 𝑂)𝐺(𝐷 − 𝑂)) = 0 ∧ ((𝐵 − 𝑂)𝐺(𝐸 − 𝑂)) = 0 ∧ ((𝐶 − 𝑂)𝐺(𝐹 − 𝑂)) = 0))    &   ⊢ (𝜑 → (((𝐴 − 𝐹)𝐺(𝐵 − 𝐹)) = 0 ∧ ((𝐵 − 𝐷)𝐺(𝐶 − 𝐷)) = 0 ∧ ((𝐶 − 𝐸)𝐺(𝐴 − 𝐸)) = 0))    &   ⊢ (𝜑 → (((𝐴 − 𝑂)𝐺(𝐵 − 𝑂)) ≠ 0 ∧ ((𝐵 − 𝑂)𝐺(𝐶 − 𝑂)) ≠ 0 ∧ ((𝐶 − 𝑂)𝐺(𝐴 − 𝑂)) ≠ 0))    ⇒   ⊢ (𝜑 → (((𝐶 − 𝑂)𝐺(𝐴 − 𝑂)) · (𝐵 − 𝐷)) = (((𝐴 − 𝑂)𝐺(𝐵 − 𝑂)) · (𝐷 − 𝐶)))
Theorem	cevath 47611*	Ceva's theorem. Let 𝐴𝐵𝐶 be a triangle and let points 𝐹, 𝐷 and 𝐸 lie on sides 𝐴𝐵, 𝐵𝐶, 𝐶𝐴 correspondingly. Suppose that cevians 𝐴𝐷, 𝐵𝐸 and 𝐶𝐹 intersect at one point 𝑂. Then triangle's sides are partitioned into segments and their lengths satisfy a certain identity. Here we obtain a bit stronger version by using complex numbers themselves instead of their absolute values. The proof goes by applying cevathlem2 47610 three times and then using cevathlem1 47609 to multiply obtained identities and prove the theorem. In the theorem statement we are using function 𝐺 as a collinearity indicator. For justification of that use, see sigarcol 47606. This is Metamath 100 proof #61. (Contributed by Saveliy Skresanov, 24-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    &   ⊢ (𝜑 → (𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ))    &   ⊢ (𝜑 → (𝐹 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐸 ∈ ℂ))    &   ⊢ (𝜑 → 𝑂 ∈ ℂ)    &   ⊢ (𝜑 → (((𝐴 − 𝑂)𝐺(𝐷 − 𝑂)) = 0 ∧ ((𝐵 − 𝑂)𝐺(𝐸 − 𝑂)) = 0 ∧ ((𝐶 − 𝑂)𝐺(𝐹 − 𝑂)) = 0))    &   ⊢ (𝜑 → (((𝐴 − 𝐹)𝐺(𝐵 − 𝐹)) = 0 ∧ ((𝐵 − 𝐷)𝐺(𝐶 − 𝐷)) = 0 ∧ ((𝐶 − 𝐸)𝐺(𝐴 − 𝐸)) = 0))    &   ⊢ (𝜑 → (((𝐴 − 𝑂)𝐺(𝐵 − 𝑂)) ≠ 0 ∧ ((𝐵 − 𝑂)𝐺(𝐶 − 𝑂)) ≠ 0 ∧ ((𝐶 − 𝑂)𝐺(𝐴 − 𝑂)) ≠ 0))    ⇒   ⊢ (𝜑 → (((𝐴 − 𝐹) · (𝐶 − 𝐸)) · (𝐵 − 𝐷)) = (((𝐹 − 𝐵) · (𝐸 − 𝐴)) · (𝐷 − 𝐶)))
21.45.2  Simple groups
Theorem	simpcntrab 47612	The center of a simple group is trivial or the group is abelian. (Contributed by SS, 3-Jan-2024.)
⊢ 𝐵 = (Base‘𝐺)    &   ⊢ 0 = (0g‘𝐺)    &   ⊢ 𝑍 = (Cntr‘𝐺)    &   ⊢ (𝜑 → 𝐺 ∈ SimpGrp)    ⇒   ⊢ (𝜑 → (𝑍 = { 0 } ∨ 𝐺 ∈ Abel))
21.46  Mathbox for Ender Ting
21.46.1  Interesting facts
Theorem	et-ltneverrefl 47613	Less-than class is never reflexive. (Contributed by Ender Ting, 22-Nov-2024.) Prefer to specify theorem domain and then apply ltnri 11325. (New usage is discouraged.)
⊢ ¬ 𝐴 < 𝐴
Theorem	et-equeucl 47614	Alternative proof that equality is left-Euclidean, using ax7 2045 directly instead of utility theorems; done for practice. (Contributed by Ender Ting, 21-Dec-2024.)
⊢ (𝑥 = 𝑧 → (𝑦 = 𝑧 → 𝑥 = 𝑦))
Theorem	et-sqrtnegnre 47615	The square root of a negative number is not a real number. (Contributed by Ender Ting, 5-Jan-2025.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐴 < 0) → ¬ (√‘𝐴) ∈ ℝ)
Theorem	quantgodel 47616	There can be no formula asserting its own non-universality, in parallel to bj-babygodel 37224; proof path is shorter but relying on a property of specialization which provability predicates do not have. For a matching proof, see quantgodelALT 47617. (Contributed by Ender Ting, 9-May-2026.)
⊢ (𝜑 ↔ ¬ ∀𝑥𝜑)    ⇒   ⊢ ⊥
Theorem	quantgodelALT 47617	There can be no formula asserting its own non-universality; follows the steps of bj-babygodel 37224. (Contributed by Ender Ting, 7-May-2026.) (New usage is discouraged.) (Proof modification is discouraged.)
⊢ (𝜑 ↔ ¬ ∀𝑥𝜑)    ⇒   ⊢ ⊥
21.46.2  Increasing sequences and subsequences
Theorem	ormklocald 47618*	If elements of a certain sequence are ordered with respect to a certain relation, then its consecutive elements satisfy that relation (so-called "local monotonicity"). (Contributed by Ender Ting, 30-Apr-2025.)
⊢ (𝜑 → 𝑅 Or 𝑆)    &   ⊢ (𝜑 → ∀𝑘 ∈ (0..^(𝑇 + 1))(𝐵‘𝑘) ∈ 𝑆)    &   ⊢ (𝜑 → ∀𝑘 ∈ (0..^𝑇)∀𝑡 ∈ (1..^(𝑇 + 1))(𝑘 < 𝑡 → (𝐵‘𝑘)𝑅(𝐵‘𝑡)))    ⇒   ⊢ (𝜑 → ∀𝑘 ∈ (0..^𝑇)(𝐵‘𝑘)𝑅(𝐵‘(𝑘 + 1)))
Theorem	ormkglobd 47619*	If all adjacent elements of a certain sequence are ordered according to a relation which is a total order on S, then any element is so related to anything to right of it (so-called "global monotonicity"). Deduction form. (Contributed by Ender Ting, 30-Apr-2025.)
⊢ (𝜑 → 𝑅 Or 𝑆)    &   ⊢ (𝜑 → ∀𝑘 ∈ (0..^(𝑇 + 1))(𝐵‘𝑘) ∈ 𝑆)    &   ⊢ (𝜑 → ∀𝑘 ∈ (0..^𝑇)(𝐵‘𝑘)𝑅(𝐵‘(𝑘 + 1)))    ⇒   ⊢ (𝜑 → ∀𝑘 ∈ (0..^𝑇)∀𝑡 ∈ (1..^(𝑇 + 1))(𝑘 < 𝑡 → (𝐵‘𝑘)𝑅(𝐵‘𝑡)))
Theorem	natlocalincr 47620*	Global monotonicity on half-open range implies local monotonicity. Inference form. (Contributed by Ender Ting, 22-Nov-2024.)
⊢ ∀𝑘 ∈ (0..^𝑇)∀𝑡 ∈ (1..^(𝑇 + 1))(𝑘 < 𝑡 → (𝐵‘𝑘) < (𝐵‘𝑡))    ⇒   ⊢ ∀𝑘 ∈ (0..^𝑇)(𝐵‘𝑘) < (𝐵‘(𝑘 + 1))
Theorem	natglobalincr 47621*	Local monotonicity on half-open integer range implies global monotonicity. Inference form. (Contributed by Ender Ting, 23-Nov-2024.)
⊢ ∀𝑘 ∈ (0..^𝑇)(𝐵‘𝑘) < (𝐵‘(𝑘 + 1))    &   ⊢ 𝑇 ∈ ℤ    ⇒   ⊢ ∀𝑘 ∈ (0..^𝑇)∀𝑡 ∈ ((𝑘 + 1)...𝑇)(𝐵‘𝑘) < (𝐵‘𝑡)
Theorem	chnsubseqword 47622	A subsequence of a chain is a word. (Contributed by Ender Ting, 22-Jan-2026.)
⊢ (𝜑 → 𝑊 ∈ ( < Chain 𝐴))    &   ⊢ (𝜑 → 𝐼 ∈ ( < Chain (0..^(♯‘𝑊))))    ⇒   ⊢ (𝜑 → (𝑊 ∘ 𝐼) ∈ Word 𝐴)
Theorem	chnsubseqwl 47623	A subsequence of a chain has the same length as its indexing sequence. (Contributed by Ender Ting, 22-Jan-2026.)
⊢ (𝜑 → 𝑊 ∈ ( < Chain 𝐴))    &   ⊢ (𝜑 → 𝐼 ∈ ( < Chain (0..^(♯‘𝑊))))    ⇒   ⊢ (𝜑 → (♯‘(𝑊 ∘ 𝐼)) = (♯‘𝐼))
Theorem	chnsubseq 47624	An order-preserving subsequence of an ordered chain is itself a chain. (Contributed by Ender Ting, 22-Jan-2026.)
⊢ (𝜑 → 𝑊 ∈ ( < Chain 𝐴))    &   ⊢ (𝜑 → 𝐼 ∈ ( < Chain (0..^(♯‘𝑊))))    &   ⊢ (𝜑 → < Po 𝐴)    ⇒   ⊢ (𝜑 → (𝑊 ∘ 𝐼) ∈ ( < Chain 𝐴))
Theorem	chnsuslle 47625	Length of a subsequence is bounded by the length of original chain. (Contributed by Ender Ting, 30-Jan-2026.)
⊢ (𝜑 → 𝑊 ∈ ( < Chain 𝐴))    &   ⊢ (𝜑 → 𝐼 ∈ ( < Chain (0..^(♯‘𝑊))))    &   ⊢ (𝜑 → < Po 𝐴)    ⇒   ⊢ (𝜑 → (♯‘(𝑊 ∘ 𝐼)) ≤ (♯‘𝑊))
Theorem	chnerlem1 47626	In a chain constructed on an equivalence relation, the last element is equivalent to any. This theorem is a translation of chnub 18684 to equivalence relations. (Contributed by Ender Ting, 29-Jan-2026.)
⊢ (𝜑 → ∼ Er 𝐴)    &   ⊢ (𝜑 → 𝐶 ∈ ( ∼ Chain 𝐴))    &   ⊢ (𝜑 → 𝐽 ∈ (0..^(♯‘𝐶)))    ⇒   ⊢ (𝜑 → (𝐶‘𝐽) ∼ (lastS‘𝐶))
Theorem	chnerlem2 47627	Lemma for chner 47629 where the I-th element comes before the J-th. (Contributed by Ender Ting, 29-Jan-2026.)
⊢ (𝜑 → ∼ Er 𝐴)    &   ⊢ (𝜑 → 𝐶 ∈ ( ∼ Chain 𝐴))    &   ⊢ (𝜑 → 𝐽 ∈ (0..^(♯‘𝐶)))    ⇒   ⊢ ((𝜑 ∧ 𝐼 ∈ (0..^𝐽)) → (𝐶‘𝐼) ∼ (𝐶‘𝐽))
Theorem	chnerlem3 47628	Lemma for chner 47629- trichotomy of integers within the word's domain. (Contributed by Ender Ting, 29-Jan-2026.)
⊢ (𝜑 → ∼ Er 𝐴)    &   ⊢ (𝜑 → 𝐶 ∈ ( ∼ Chain 𝐴))    &   ⊢ (𝜑 → 𝐽 ∈ (0..^(♯‘𝐶)))    &   ⊢ (𝜑 → 𝐼 ∈ (0..^(♯‘𝐶)))    ⇒   ⊢ (𝜑 → (𝐼 ∈ (0..^𝐽) ∨ 𝐽 ∈ (0..^𝐼) ∨ 𝐼 = 𝐽))
Theorem	chner 47629	Any two elements are equivalent in a chain constructed on an equivalence relation. (Contributed by Ender Ting, 29-Jan-2026.)
⊢ (𝜑 → ∼ Er 𝐴)    &   ⊢ (𝜑 → 𝐶 ∈ ( ∼ Chain 𝐴))    &   ⊢ (𝜑 → 𝐽 ∈ (0..^(♯‘𝐶)))    &   ⊢ (𝜑 → 𝐼 ∈ (0..^(♯‘𝐶)))    ⇒   ⊢ (𝜑 → (𝐶‘𝐼) ∼ (𝐶‘𝐽))
21.46.3  Scratchpad for number theory
Theorem	evenwodadd 47630	If an integer is multiplied by its sum with an odd number (thus changing its parity), the result is even. (Contributed by Ender Ting, 30-Apr-2025.)
⊢ (𝜑 → 𝑖 ∈ ℤ)    &   ⊢ (𝜑 → 𝑗 ∈ ℤ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑗)    ⇒   ⊢ (𝜑 → 2 ∥ (𝑖 · (𝑖 + 𝑗)))
21.46.4  Scratchpad for math on real numbers
Theorem	squeezedltsq 47631	If a real value is squeezed between two others, its square is less than square of at least one of them. Deduction form. (Contributed by Ender Ting, 31-Oct-2025.)
⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ)    &   ⊢ (𝜑 → 𝐶 ∈ ℝ)    &   ⊢ (𝜑 → 𝐴 < 𝐵)    &   ⊢ (𝜑 → 𝐵 < 𝐶)    ⇒   ⊢ (𝜑 → ((𝐵 · 𝐵) < (𝐴 · 𝐴) ∨ (𝐵 · 𝐵) < (𝐶 · 𝐶)))
Theorem	sqrtnnaa 47632	Square root of a natural number is algebraic. (Contributed by Ender Ting, 21-Jul-2026.)
⊢ (𝐴 ∈ ℕ → (√‘𝐴) ∈ 𝔸)
Theorem	sqrtnzqaa 47633	Square root of a nonzero rational is algebraic. (Contributed by Ender Ting, 22-Jul-2026.)
⊢ ((𝐴 ∈ ℚ ∧ 𝐴 ≠ 0) → (√‘𝐴) ∈ 𝔸)
Theorem	sqrtqaa 47634	Square root of a rational number is algebraic. (Contributed by Ender Ting, 22-Jul-2026.)
⊢ (𝐴 ∈ ℚ → (√‘𝐴) ∈ 𝔸)
Theorem	nthrucw 47635*	Some number sets form a chain of proper subsets. This is rephrasing nthruc 16314 as a statement about chains; the hypothesis sets the ordering relation to be "is a proper subset". The theorem talks about singleton 1, natural numbers, natural-or-zero numbers, integers, rational numbers, algebraic reals (the definition includes complex numbers as algebraic so intersection is taken), real numbers and complex numbers, which are proper subsets in order. (Contributed by Ender Ting, 29-Jan-2026.)
⊢ < = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ⊊ 𝑦}    ⇒   ⊢ ⟨“{1}ℕℕ0ℤℚ(𝔸 ∩ ℝ)ℝℂ”⟩ ∈ ( < Chain V)
Theorem	sin3t 47636	Triple-angle formula for sine, in pure sine form. (Contributed by Ender Ting, 16-Mar-2026.)
⊢ (𝐴 ∈ ℂ → (sin‘(3 · 𝐴)) = ((3 · (sin‘𝐴)) − (4 · ((sin‘𝐴)↑3))))
Theorem	cos3t 47637	Triple-angle formula for cosine, in pure cosine form. (Contributed by Ender Ting, 16-Mar-2026.)
⊢ (𝐴 ∈ ℂ → (cos‘(3 · 𝐴)) = ((4 · ((cos‘𝐴)↑3)) − (3 · (cos‘𝐴))))
Theorem	sin5tlem1 47638	Lemma 1 for quintupled angle sine calculation, expanding triple-angle sine times double-angle cosine. (Contributed by Ender Ting, 16-Mar-2026.)
⊢ (𝑁 ∈ ℂ → (((3 · 𝑁) − (4 · (𝑁↑3))) · (1 − (2 · (𝑁↑2)))) = (((8 · (𝑁↑5)) − (;10 · (𝑁↑3))) + (3 · 𝑁)))
Theorem	sin5tlem2 47639	Lemma 2 for quintupled angle sine calculation, multiplicating triple angle cosine by cosine straight and converting into sine. (Contributed by Ender Ting, 16-Apr-2026.)
⊢ ((𝑁 ∈ ℂ ∧ 𝑀 ∈ ℂ ∧ (𝑁↑2) = (1 − (𝑀↑2))) → (((4 · (𝑁↑3)) − (3 · 𝑁)) · 𝑁) = ((4 · ((1 − (2 · (𝑀↑2))) + (𝑀↑4))) − (3 · (1 − (𝑀↑2)))))
Theorem	sin5tlem3 47640	Lemma 3 for quintupled angle sine calculation, multiplicating triple angle cosine by double angle sine. (Contributed by Ender Ting, 16-Apr-2026.)
⊢ ((𝑁 ∈ ℂ ∧ 𝑀 ∈ ℂ ∧ (𝑁↑2) = (1 − (𝑀↑2))) → (((4 · (𝑁↑3)) − (3 · 𝑁)) · (2 · (𝑀 · 𝑁))) = (((4 · ((1 − (2 · (𝑀↑2))) + (𝑀↑4))) − (3 · (1 − (𝑀↑2)))) · (2 · 𝑀)))
Theorem	sin5tlem4 47641	Lemma 4 for quintupled angle sine calculation: expanding lemma 3 result to difference of polynomials. (Contributed by Ender Ting, 17-Apr-2026.)
⊢ ((𝑁 ∈ ℂ ∧ 𝑀 ∈ ℂ ∧ (𝑁↑2) = (1 − (𝑀↑2))) → (((4 · (𝑁↑3)) − (3 · 𝑁)) · (2 · (𝑀 · 𝑁))) = ((((8 · (𝑀↑5)) − (;16 · (𝑀↑3))) + (8 · 𝑀)) − ((6 · 𝑀) − (6 · (𝑀↑3)))))
Theorem	sin5tlem5 47642	Lemma 5 for quintupled angle sine calculation: sine of triple-angle and double-angle sum, as a polynomial in sine straight. (Contributed by Ender Ting, 17-Apr-2026.)
⊢ ((𝑁 ∈ ℂ ∧ 𝑀 ∈ ℂ ∧ (𝑁↑2) = (1 − (𝑀↑2))) → ((((3 · 𝑀) − (4 · (𝑀↑3))) · (1 − (2 · (𝑀↑2)))) + (((4 · (𝑁↑3)) − (3 · 𝑁)) · (2 · (𝑀 · 𝑁)))) = (((;16 · (𝑀↑5)) − (;20 · (𝑀↑3))) + (5 · 𝑀)))
Theorem	sin5t 47643	Five-times-angle formula for sine, in pure sine form. (Contributed by Ender Ting, 17-Apr-2026.)
⊢ (𝐴 ∈ ℂ → (sin‘(5 · 𝐴)) = (((;16 · ((sin‘𝐴)↑5)) − (;20 · ((sin‘𝐴)↑3))) + (5 · (sin‘𝐴))))
Theorem	cos5t 47644	Five-times-angle formula for cosine, in pure cosine form. (Contributed by Ender Ting, 20-Apr-2026.)
⊢ (𝐴 ∈ ℂ → (cos‘(5 · 𝐴)) = (((;16 · ((cos‘𝐴)↑5)) − (;20 · ((cos‘𝐴)↑3))) + (5 · (cos‘𝐴))))
Theorem	cos5teq 47645	Five-times-angle formula for cosine, substitution helper. (Contributed by Ender Ting, 9-May-2026.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 = (5 · 𝐴) ∧ 𝐶 = (cos‘𝐴)) → (cos‘𝐵) = (((;16 · (𝐶↑5)) − (;20 · (𝐶↑3))) + (5 · 𝐶)))
Theorem	goldrarr 47646	The golden ratio is a real value. (Contributed by Ender Ting, 15-Mar-2026.)
⊢ 𝐹 = (2 · (cos‘(π / 5)))    ⇒   ⊢ 𝐹 ∈ ℝ
Theorem	goldrasin 47647	Alternative trigonometric formula for the golden ratio. (Contributed by Ender Ting, 15-Mar-2026.)
⊢ 𝐹 = (2 · (cos‘(π / 5)))    ⇒   ⊢ 𝐹 = (2 · (sin‘(π · (3 / ;10))))
Theorem	goldrapos 47648	Golden ratio is positive. (Contributed by Ender Ting, 16-Apr-2026.)
⊢ 𝐹 = (2 · (cos‘(π / 5)))    ⇒   ⊢ 0 < 𝐹
Theorem	goldrarp 47649	The golden ratio is a positive real. (Contributed by Ender Ting, 16-Apr-2026.)
⊢ 𝐹 = (2 · (cos‘(π / 5)))    ⇒   ⊢ 𝐹 ∈ ℝ+
Theorem	goldracos5teq 47650	Lemma 1 for determining the value of golden ratio. (Contributed by Ender Ting, 9-May-2026.)
⊢ 𝐹 = (2 · (cos‘(π / 5)))    ⇒   ⊢ (cos‘π) = (((;16 · ((𝐹 / 2)↑5)) − (;20 · ((𝐹 / 2)↑3))) + (5 · (𝐹 / 2)))
Theorem	goldratmolem2 47651	Lemma 2 for determining the value of golden ratio. (Contributed by Ender Ting, 9-May-2026.)
⊢ 𝐹 = (2 · (cos‘(π / 5)))    ⇒   ⊢ -1 = ((((𝐹↑5) / 2) − (5 · ((𝐹↑3) / 2))) + (5 · (𝐹 / 2)))
Theorem	lambert0 47652	A value of Lambert W (product logarithm) function at zero. (Contributed by Ender Ting, 13-Nov-2025.)
⊢ 𝑅 = ◡(𝑥 ∈ ℂ ↦ (𝑥 · (exp‘𝑥)))    ⇒   ⊢ 0𝑅0
Theorem	lamberte 47653	A value of Lambert W (product logarithm) function at e. (Contributed by Ender Ting, 13-Nov-2025.)
⊢ 𝑅 = ◡(𝑥 ∈ ℂ ↦ (𝑥 · (exp‘𝑥)))    ⇒   ⊢ e𝑅1
Theorem	cjnpoly 47654	Complex conjugation operator is not a polynomial with complex coefficients. Indeed; if it was, then multiplying 𝑥 conjugate by 𝑥 itself and adding 1 would yield a nowhere-zero non-constant polynomial, contrary to the fta 27255. (Contributed by Ender Ting, 8-Dec-2025.)
⊢ ¬ ∗ ∈ (Poly‘ℂ)
Theorem	tannpoly 47655	The tangent function is not a polynomial with complex coefficients, as it is not defined on the whole complex plane. (Contributed by Ender Ting, 10-Dec-2025.)
⊢ ¬ tan ∈ (Poly‘ℂ)
Theorem	sinnpoly 47656	Sine function is not a polynomial with complex coefficients. Indeed, it has infinitely many zeros but is not constant zero, contrary to fta1 26480. (Contributed by Ender Ting, 10-Dec-2025.)
⊢ ¬ sin ∈ (Poly‘ℂ)
21.47  Mathbox for Jarvin Udandy
Theorem	hirstL-ax3 47657	The third axiom of a system called "L" but proven to be a theorem since set.mm uses a different third axiom. This is named hirst after Holly P. Hirst and Jeffry L. Hirst. Axiom A3 of [Mendelson] p. 35. (Contributed by Jarvin Udandy, 7-Feb-2015.) (Proof modification is discouraged.)
⊢ ((¬ 𝜑 → ¬ 𝜓) → ((¬ 𝜑 → 𝜓) → 𝜑))
Theorem	ax3h 47658	Recover ax-3 8 from hirstL-ax3 47657. (Contributed by Jarvin Udandy, 3-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ ((¬ 𝜑 → ¬ 𝜓) → (𝜓 → 𝜑))
Theorem	aibandbiaiffaiffb 47659	A closed form showing (a implies b and b implies a) same-as (a same-as b). (Contributed by Jarvin Udandy, 3-Sep-2016.)
⊢ (((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) ↔ (𝜑 ↔ 𝜓))
Theorem	aibandbiaiaiffb 47660	A closed form showing (a implies b and b implies a) implies (a same-as b). (Contributed by Jarvin Udandy, 3-Sep-2016.)
⊢ (((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) → (𝜑 ↔ 𝜓))
Theorem	notatnand 47661	Do not use. Use intnanr instead. Given not a, there exists a proof for not (a and b). (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ¬ 𝜑    ⇒   ⊢ ¬ (𝜑 ∧ 𝜓)
Theorem	aistia 47662	Given a is equivalent to ⊤, there exists a proof for a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
⊢ (𝜑 ↔ ⊤)    ⇒   ⊢ 𝜑
Theorem	aisfina 47663	Given a is equivalent to ⊥, there exists a proof for not a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
⊢ (𝜑 ↔ ⊥)    ⇒   ⊢ ¬ 𝜑
Theorem	bothtbothsame 47664	Given both a, b are equivalent to ⊤, there exists a proof for a is the same as b. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ (𝜑 ↔ ⊤)    &   ⊢ (𝜓 ↔ ⊤)    ⇒   ⊢ (𝜑 ↔ 𝜓)
Theorem	bothfbothsame 47665	Given both a, b are equivalent to ⊥, there exists a proof for a is the same as b. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊥)    ⇒   ⊢ (𝜑 ↔ 𝜓)
Theorem	aiffbbtat 47666	Given a is equivalent to b, b is equivalent to ⊤ there exists a proof for a is equivalent to T. (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ (𝜑 ↔ 𝜓)    &   ⊢ (𝜓 ↔ ⊤)    ⇒   ⊢ (𝜑 ↔ ⊤)
Theorem	aisbbisfaisf 47667	Given a is equivalent to b, b is equivalent to ⊥ there exists a proof for a is equivalent to F. (Contributed by Jarvin Udandy, 30-Aug-2016.)
⊢ (𝜑 ↔ 𝜓)    &   ⊢ (𝜓 ↔ ⊥)    ⇒   ⊢ (𝜑 ↔ ⊥)
Theorem	axorbtnotaiffb 47668	Given a is exclusive to b, there exists a proof for (not (a if-and-only-if b)); df-xor 1541 is a closed form of this. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜓)    ⇒   ⊢ ¬ (𝜑 ↔ 𝜓)
Theorem	aiffnbandciffatnotciffb 47669	Given a is equivalent to (not b), c is equivalent to a, there exists a proof for ( not ( c iff b ) ). (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ ¬ 𝜓)    &   ⊢ (𝜒 ↔ 𝜑)    ⇒   ⊢ ¬ (𝜒 ↔ 𝜓)
Theorem	axorbciffatcxorb 47670	Given a is equivalent to (not b), c is equivalent to a. there exists a proof for ( c xor b ). (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜓)    &   ⊢ (𝜒 ↔ 𝜑)    ⇒   ⊢ (𝜒 ⊻ 𝜓)
Theorem	aibnbna 47671	Given a implies b, (not b), there exists a proof for (not a). (Contributed by Jarvin Udandy, 1-Sep-2016.)
⊢ (𝜑 → 𝜓)    &   ⊢ ¬ 𝜓    ⇒   ⊢ ¬ 𝜑
Theorem	aibnbaif 47672	Given a implies b, not b, there exists a proof for a is F. (Contributed by Jarvin Udandy, 1-Sep-2016.)
⊢ (𝜑 → 𝜓)    &   ⊢ ¬ 𝜓    ⇒   ⊢ (𝜑 ↔ ⊥)
Theorem	aiffbtbat 47673	Given a is equivalent to b, T. is equivalent to b. there exists a proof for a is equivalent to T. (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ (𝜑 ↔ 𝜓)    &   ⊢ (⊤ ↔ 𝜓)    ⇒   ⊢ (𝜑 ↔ ⊤)
Theorem	astbstanbst 47674	Given a is equivalent to T., also given that b is equivalent to T, there exists a proof for a and b is equivalent to T. (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ (𝜑 ↔ ⊤)    &   ⊢ (𝜓 ↔ ⊤)    ⇒   ⊢ ((𝜑 ∧ 𝜓) ↔ ⊤)
Theorem	aistbistaandb 47675	Given a is equivalent to T., also given that b is equivalent to T, there exists a proof for (a and b). (Contributed by Jarvin Udandy, 9-Sep-2016.)
⊢ (𝜑 ↔ ⊤)    &   ⊢ (𝜓 ↔ ⊤)    ⇒   ⊢ (𝜑 ∧ 𝜓)
Theorem	aisbnaxb 47676	Given a is equivalent to b, there exists a proof for (not (a xor b)). (Contributed by Jarvin Udandy, 28-Aug-2016.)
⊢ (𝜑 ↔ 𝜓)    ⇒   ⊢ ¬ (𝜑 ⊻ 𝜓)
Theorem	atbiffatnnb 47677	If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.)
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))
Theorem	bisaiaisb 47678	Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ((𝜓 ↔ 𝜑) → (𝜑 ↔ 𝜓))
Theorem	atbiffatnnbalt 47679	If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))
Theorem	abnotbtaxb 47680	Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ 𝜑    &   ⊢ ¬ 𝜓    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	abnotataxb 47681	Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ¬ 𝜑    &   ⊢ 𝜓    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	conimpf 47682	Assuming a, not b, and a implies b, there exists a proof that a is false.) (Contributed by Jarvin Udandy, 28-Aug-2016.)
⊢ 𝜑    &   ⊢ ¬ 𝜓    &   ⊢ (𝜑 → 𝜓)    ⇒   ⊢ (𝜑 ↔ ⊥)
Theorem	conimpfalt 47683	Assuming a, not b, and a implies b, there exists a proof that a is false.) (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ 𝜑    &   ⊢ ¬ 𝜓    &   ⊢ (𝜑 → 𝜓)    ⇒   ⊢ (𝜑 ↔ ⊥)
Theorem	aistbisfiaxb 47684	Given a is equivalent to T., Given b is equivalent to F. there exists a proof for a-xor-b. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ (𝜑 ↔ ⊤)    &   ⊢ (𝜓 ↔ ⊥)    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	aisfbistiaxb 47685	Given a is equivalent to F., Given b is equivalent to T., there exists a proof for a-xor-b. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	aifftbifffaibif 47686	Given a is equivalent to T., Given b is equivalent to F., there exists a proof for that a implies b is false. (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ (𝜑 ↔ ⊤)    &   ⊢ (𝜓 ↔ ⊥)    ⇒   ⊢ ((𝜑 → 𝜓) ↔ ⊥)
Theorem	aifftbifffaibifff 47687	Given a is equivalent to T., Given b is equivalent to F., there exists a proof for that a iff b is false. (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ (𝜑 ↔ ⊤)    &   ⊢ (𝜓 ↔ ⊥)    ⇒   ⊢ ((𝜑 ↔ 𝜓) ↔ ⊥)
Theorem	atnaiana 47688	Given a, it is not the case a implies a self contradiction. (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ 𝜑    ⇒   ⊢ ¬ (𝜑 → (𝜑 ∧ ¬ 𝜑))
Theorem	ainaiaandna 47689	Given a, a implies it is not the case a implies a self contradiction. (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ 𝜑    ⇒   ⊢ (𝜑 → ¬ (𝜑 → (𝜑 ∧ ¬ 𝜑)))
Theorem	abcdta 47690	Given (((a and b) and c) and d), there exists a proof for a. (Contributed by Jarvin Udandy, 3-Sep-2016.)
⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ∧ 𝜃)    ⇒   ⊢ 𝜑
Theorem	abcdtb 47691	Given (((a and b) and c) and d), there exists a proof for b. (Contributed by Jarvin Udandy, 3-Sep-2016.)
⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ∧ 𝜃)    ⇒   ⊢ 𝜓
Theorem	abcdtc 47692	Given (((a and b) and c) and d), there exists a proof for c. (Contributed by Jarvin Udandy, 3-Sep-2016.)
⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ∧ 𝜃)    ⇒   ⊢ 𝜒
Theorem	abcdtd 47693	Given (((a and b) and c) and d), there exists a proof for d. (Contributed by Jarvin Udandy, 3-Sep-2016.)
⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ∧ 𝜃)    ⇒   ⊢ 𝜃
Theorem	abciffcbatnabciffncba 47694	Operands in a biconditional expression converted negated. Additionally biconditional converted to show antecedent implies sequent. Closed form. (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ (¬ ((𝜑 ∧ 𝜓) ∧ 𝜒) → ¬ ((𝜒 ∧ 𝜓) ∧ 𝜑))
Theorem	abciffcbatnabciffncbai 47695	Operands in a biconditional expression converted negated. Additionally biconditional converted to show antecedent implies sequent. (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ((𝜒 ∧ 𝜓) ∧ 𝜑))    ⇒   ⊢ (¬ ((𝜑 ∧ 𝜓) ∧ 𝜒) → ¬ ((𝜒 ∧ 𝜓) ∧ 𝜑))
Theorem	nabctnabc 47696	not ( a -> ( b /\ c ) ) we can show: not a implies ( b /\ c ). (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ ¬ (𝜑 → (𝜓 ∧ 𝜒))    ⇒   ⊢ (¬ 𝜑 → (𝜓 ∧ 𝜒))
Theorem	jabtaib 47697	For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.)
⊢ (𝜑 ∧ 𝜓)    ⇒   ⊢ (𝜑 → 𝜓)
Theorem	onenotinotbothi 47698	From one negated implication it is not the case its nonnegated form and a random others are both true. (Contributed by Jarvin Udandy, 11-Sep-2020.)
⊢ ¬ (𝜑 → 𝜓)    ⇒   ⊢ ¬ ((𝜑 → 𝜓) ∧ (𝜒 → 𝜃))
Theorem	twonotinotbothi 47699	From these two negated implications it is not the case their nonnegated forms are both true. (Contributed by Jarvin Udandy, 11-Sep-2020.)
⊢ ¬ (𝜑 → 𝜓)    &   ⊢ ¬ (𝜒 → 𝜃)    ⇒   ⊢ ¬ ((𝜑 → 𝜓) ∧ (𝜒 → 𝜃))
Theorem	clifte 47700	show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
⊢ (𝜑 ∧ ¬ 𝜒)    &   ⊢ 𝜃    ⇒   ⊢ (𝜃 ↔ ((𝜑 ∧ ¬ 𝜒) ∨ (𝜓 ∧ 𝜒)))