P. 474 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 47301-47400   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	sigarid 47301*	Signed area of a flat parallelogram is zero. (Contributed by Saveliy Skresanov, 20-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    ⇒   ⊢ (𝐴 ∈ ℂ → (𝐴𝐺𝐴) = 0)
Theorem	sigarexp 47302*	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 47303*	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 47304*	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 47305*	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 47306*	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 47307*	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 47308*	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 47309*	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 47310	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 47311*	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 47312*	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 47311 three times and then using cevathlem1 47310 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 47307. 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)))
21.44.2  Simple groups
Theorem	simpcntrab 47313	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))
21.45  Mathbox for Ender Ting
21.45.1  Increasing sequences and subsequences
Theorem	et-ltneverrefl 47314	Less-than class is never reflexive. (Contributed by Ender Ting, 22-Nov-2024.) Prefer to specify theorem domain and then apply ltnri 11244. (New usage is discouraged.)
⊢ ¬ 𝐴 < 𝐴
Theorem	et-equeucl 47315	Alternative proof that equality is left-Euclidean, using ax7 2018 directly instead of utility theorems; done for practice. (Contributed by Ender Ting, 21-Dec-2024.)
⊢ (𝑥 = 𝑧 → (𝑦 = 𝑧 → 𝑥 = 𝑦))
Theorem	et-sqrtnegnre 47316	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	ormklocald 47317*	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 47318*	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 47319*	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 47320*	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 47321	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 47322	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 47323	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 47324	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 47325	In a chain constructed on an equivalence relation, the last element is equivalent to any. This theorem is a translation of chnub 18577 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 47326	Lemma for chner 47328 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 47327	Lemma for chner 47328- 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 47328	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))
Theorem	nthrucw 47329*	Some number sets form a chain of proper subsets. This is rephrasing nthruc 16208 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)
21.45.2  Scratchpad for number theory
Theorem	evenwodadd 47330	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)))
21.45.3  Scratchpad for math on real numbers
Theorem	squeezedltsq 47331	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	lambert0 47332	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 47333	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 47334	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 27061. (Contributed by Ender Ting, 8-Dec-2025.)
⊢ ¬ ∗ ∈ (Poly‘ℂ)
Theorem	tannpoly 47335	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 47336	Sine function is not a polynomial with complex coefficients. Indeed, it has infinitely many zeros but is not constant zero, contrary to fta1 26287. (Contributed by Ender Ting, 10-Dec-2025.)
⊢ ¬ sin ∈ (Poly‘ℂ)
21.46  Mathbox for Jarvin Udandy
Theorem	hirstL-ax3 47337	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 47338	Recover ax-3 8 from hirstL-ax3 47337. (Contributed by Jarvin Udandy, 3-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ ((¬ 𝜑 → ¬ 𝜓) → (𝜓 → 𝜑))
Theorem	aibandbiaiffaiffb 47339	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 47340	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 47341	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 47342	Given a is equivalent to ⊤, there exists a proof for a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
⊢ (𝜑 ↔ ⊤)    ⇒   ⊢ 𝜑
Theorem	aisfina 47343	Given a is equivalent to ⊥, there exists a proof for not a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
⊢ (𝜑 ↔ ⊥)    ⇒   ⊢ ¬ 𝜑
Theorem	bothtbothsame 47344	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 47345	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 47346	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 47347	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 47348	Given a is exclusive to b, there exists a proof for (not (a if-and-only-if b)); df-xor 1514 is a closed form of this. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜓)    ⇒   ⊢ ¬ (𝜑 ↔ 𝜓)
Theorem	aiffnbandciffatnotciffb 47349	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 47350	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 47351	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 47352	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 47353	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 47354	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 47355	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 47356	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 47357	If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.)
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))
Theorem	bisaiaisb 47358	Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ((𝜓 ↔ 𝜑) → (𝜑 ↔ 𝜓))
Theorem	atbiffatnnbalt 47359	If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))
Theorem	abnotbtaxb 47360	Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ 𝜑    &   ⊢ ¬ 𝜓    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	abnotataxb 47361	Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ¬ 𝜑    &   ⊢ 𝜓    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	conimpf 47362	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 47363	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 47364	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 47365	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 47366	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 47367	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 47368	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 47369	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 47370	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 47371	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 47372	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 47373	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 47374	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 47375	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 47376	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 47377	For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.)
⊢ (𝜑 ∧ 𝜓)    ⇒   ⊢ (𝜑 → 𝜓)
Theorem	onenotinotbothi 47378	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 47379	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 47380	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)))
Theorem	cliftet 47381	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)))
Theorem	clifteta 47382	show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
⊢ ((𝜑 ∧ ¬ 𝜒) ∨ (𝜓 ∧ 𝜒))    &   ⊢ 𝜃    ⇒   ⊢ (𝜃 ↔ ((𝜑 ∧ ¬ 𝜒) ∨ (𝜓 ∧ 𝜒)))
Theorem	cliftetb 47383	show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
⊢ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ ¬ 𝜒))    &   ⊢ 𝜃    ⇒   ⊢ (𝜃 ↔ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ ¬ 𝜒)))
Theorem	confun 47384	Given the hypotheses there exists a proof for (c implies ( d iff a ) ). (Contributed by Jarvin Udandy, 6-Sep-2020.)
⊢ 𝜑    &   ⊢ (𝜒 → 𝜓)    &   ⊢ (𝜒 → 𝜃)    &   ⊢ (𝜑 → (𝜑 → 𝜓))    ⇒   ⊢ (𝜒 → (𝜃 ↔ 𝜑))
Theorem	confun2 47385	Confun simplified to two propositions. (Contributed by Jarvin Udandy, 6-Sep-2020.)
⊢ (𝜓 → 𝜑)    &   ⊢ (𝜓 → ¬ (𝜓 → (𝜓 ∧ ¬ 𝜓)))    &   ⊢ ((𝜓 → 𝜑) → ((𝜓 → 𝜑) → 𝜑))    ⇒   ⊢ (𝜓 → (¬ (𝜓 → (𝜓 ∧ ¬ 𝜓)) ↔ (𝜓 → 𝜑)))
Theorem	confun3 47386	Confun's more complex form where both a,d have been "defined". (Contributed by Jarvin Udandy, 6-Sep-2020.)
⊢ (𝜑 ↔ (𝜒 → 𝜓))    &   ⊢ (𝜃 ↔ ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   ⊢ (𝜒 → 𝜓)    &   ⊢ (𝜒 → ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   ⊢ ((𝜒 → 𝜓) → ((𝜒 → 𝜓) → 𝜓))    ⇒   ⊢ (𝜒 → (¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)) ↔ (𝜒 → 𝜓)))
Theorem	confun4 47387	An attempt at derivative. Resisted simplest path to a proof. (Contributed by Jarvin Udandy, 6-Sep-2020.)
⊢ 𝜑    &   ⊢ ((𝜑 → 𝜓) → 𝜓)    &   ⊢ (𝜓 → (𝜑 → 𝜒))    &   ⊢ ((𝜒 → 𝜃) → ((𝜑 → 𝜃) ↔ 𝜓))    &   ⊢ (𝜏 ↔ (𝜒 → 𝜃))    &   ⊢ (𝜂 ↔ ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   ⊢ 𝜓    &   ⊢ (𝜒 → 𝜃)    ⇒   ⊢ (𝜒 → (𝜓 → 𝜏))
Theorem	confun5 47388	An attempt at derivative. Resisted simplest path to a proof. Interesting that ch, th, ta, et were all provable. (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ 𝜑    &   ⊢ ((𝜑 → 𝜓) → 𝜓)    &   ⊢ (𝜓 → (𝜑 → 𝜒))    &   ⊢ ((𝜒 → 𝜃) → ((𝜑 → 𝜃) ↔ 𝜓))    &   ⊢ (𝜏 ↔ (𝜒 → 𝜃))    &   ⊢ (𝜂 ↔ ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   ⊢ 𝜓    &   ⊢ (𝜒 → 𝜃)    ⇒   ⊢ (𝜒 → (𝜂 ↔ 𝜏))
Theorem	plcofph 47389	Given, a,b and a "definition" for c, c is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
⊢ (𝜒 ↔ ((((𝜑 ∧ 𝜓) ↔ 𝜑) → (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))) ∧ (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))))    &   ⊢ 𝜑    &   ⊢ 𝜓    ⇒   ⊢ 𝜒
Theorem	pldofph 47390	Given, a,b c, d, "definition" for e, e is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
⊢ (𝜏 ↔ ((𝜒 → 𝜃) ∧ (𝜑 ↔ 𝜒) ∧ ((𝜑 → 𝜓) → (𝜓 ↔ 𝜃))))    &   ⊢ 𝜑    &   ⊢ 𝜓    &   ⊢ 𝜒    &   ⊢ 𝜃    ⇒   ⊢ 𝜏
Theorem	plvcofph 47391	Given, a,b,d, and "definitions" for c, e, f: f is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
⊢ (𝜒 ↔ ((((𝜑 ∧ 𝜓) ↔ 𝜑) → (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))) ∧ (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))))    &   ⊢ (𝜏 ↔ ((𝜒 → 𝜃) ∧ (𝜑 ↔ 𝜒) ∧ ((𝜑 → 𝜓) → (𝜓 ↔ 𝜃))))    &   ⊢ (𝜂 ↔ (𝜒 ∧ 𝜏))    &   ⊢ 𝜑    &   ⊢ 𝜓    &   ⊢ 𝜃    ⇒   ⊢ 𝜂
Theorem	plvcofphax 47392	Given, a,b,d, and "definitions" for c, e, f, g: g is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
⊢ (𝜒 ↔ ((((𝜑 ∧ 𝜓) ↔ 𝜑) → (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))) ∧ (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))))    &   ⊢ (𝜏 ↔ ((𝜒 → 𝜃) ∧ (𝜑 ↔ 𝜒) ∧ ((𝜑 → 𝜓) → (𝜓 ↔ 𝜃))))    &   ⊢ (𝜂 ↔ (𝜒 ∧ 𝜏))    &   ⊢ 𝜑    &   ⊢ 𝜓    &   ⊢ 𝜃    &   ⊢ (𝜁 ↔ ¬ (𝜓 ∧ ¬ 𝜏))    ⇒   ⊢ 𝜁
Theorem	plvofpos 47393	rh is derivable because ONLY one of ch, th, ta, et is implied by mu. (Contributed by Jarvin Udandy, 11-Sep-2020.)
⊢ (𝜒 ↔ (¬ 𝜑 ∧ ¬ 𝜓))    &   ⊢ (𝜃 ↔ (¬ 𝜑 ∧ 𝜓))    &   ⊢ (𝜏 ↔ (𝜑 ∧ ¬ 𝜓))    &   ⊢ (𝜂 ↔ (𝜑 ∧ 𝜓))    &   ⊢ (𝜁 ↔ (((((¬ ((𝜇 → 𝜒) ∧ (𝜇 → 𝜃)) ∧ ¬ ((𝜇 → 𝜒) ∧ (𝜇 → 𝜏))) ∧ ¬ ((𝜇 → 𝜒) ∧ (𝜒 → 𝜂))) ∧ ¬ ((𝜇 → 𝜃) ∧ (𝜇 → 𝜏))) ∧ ¬ ((𝜇 → 𝜃) ∧ (𝜇 → 𝜂))) ∧ ¬ ((𝜇 → 𝜏) ∧ (𝜇 → 𝜂))))    &   ⊢ (𝜎 ↔ (((𝜇 → 𝜒) ∨ (𝜇 → 𝜃)) ∨ ((𝜇 → 𝜏) ∨ (𝜇 → 𝜂))))    &   ⊢ (𝜌 ↔ (𝜁 ∧ 𝜎))    &   ⊢ 𝜁    &   ⊢ 𝜎    ⇒   ⊢ 𝜌
Theorem	mdandyv0 47394	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊥)    &   ⊢ (𝜃 ↔ ⊥)    &   ⊢ (𝜏 ↔ ⊥)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜑) ∧ (𝜃 ↔ 𝜑)) ∧ (𝜏 ↔ 𝜑)) ∧ (𝜂 ↔ 𝜑))
Theorem	mdandyv1 47395	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊤)    &   ⊢ (𝜃 ↔ ⊥)    &   ⊢ (𝜏 ↔ ⊥)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜓) ∧ (𝜃 ↔ 𝜑)) ∧ (𝜏 ↔ 𝜑)) ∧ (𝜂 ↔ 𝜑))
Theorem	mdandyv2 47396	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊥)    &   ⊢ (𝜃 ↔ ⊤)    &   ⊢ (𝜏 ↔ ⊥)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜑) ∧ (𝜃 ↔ 𝜓)) ∧ (𝜏 ↔ 𝜑)) ∧ (𝜂 ↔ 𝜑))
Theorem	mdandyv3 47397	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊤)    &   ⊢ (𝜃 ↔ ⊤)    &   ⊢ (𝜏 ↔ ⊥)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜓) ∧ (𝜃 ↔ 𝜓)) ∧ (𝜏 ↔ 𝜑)) ∧ (𝜂 ↔ 𝜑))
Theorem	mdandyv4 47398	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊥)    &   ⊢ (𝜃 ↔ ⊥)    &   ⊢ (𝜏 ↔ ⊤)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜑) ∧ (𝜃 ↔ 𝜑)) ∧ (𝜏 ↔ 𝜓)) ∧ (𝜂 ↔ 𝜑))
Theorem	mdandyv5 47399	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊤)    &   ⊢ (𝜃 ↔ ⊥)    &   ⊢ (𝜏 ↔ ⊤)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜓) ∧ (𝜃 ↔ 𝜑)) ∧ (𝜏 ↔ 𝜓)) ∧ (𝜂 ↔ 𝜑))
Theorem	mdandyv6 47400	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊥)    &   ⊢ (𝜃 ↔ ⊤)    &   ⊢ (𝜏 ↔ ⊤)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜑) ∧ (𝜃 ↔ 𝜓)) ∧ (𝜏 ↔ 𝜓)) ∧ (𝜂 ↔ 𝜑))