P. 464 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 46301-46400   *Has distinct variable group(s)

Type	Label	Description
Statement
21.41  Mathbox for Saveliy Skresanov
21.41.1  Ceva's theorem
Theorem	sigarval 46301*	Define the signed area by treating complex numbers as vectors with two components. (Contributed by Saveliy Skresanov, 19-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴𝐺𝐵) = (ℑ‘((∗‘𝐴) · 𝐵)))
Theorem	sigarim 46302*	Signed area takes value in reals. (Contributed by Saveliy Skresanov, 19-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴𝐺𝐵) ∈ ℝ)
Theorem	sigarac 46303*	Signed area is anticommutative. (Contributed by Saveliy Skresanov, 19-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴𝐺𝐵) = -(𝐵𝐺𝐴))
Theorem	sigaraf 46304*	Signed area is additive by the first argument. (Contributed by Saveliy Skresanov, 19-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ((𝐴 + 𝐶)𝐺𝐵) = ((𝐴𝐺𝐵) + (𝐶𝐺𝐵)))
Theorem	sigarmf 46305*	Signed area is additive (with respect to subtraction) by the first argument. (Contributed by Saveliy Skresanov, 19-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ((𝐴 − 𝐶)𝐺𝐵) = ((𝐴𝐺𝐵) − (𝐶𝐺𝐵)))
Theorem	sigaras 46306*	Signed area is additive by the second argument. (Contributed by Saveliy Skresanov, 19-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (𝐴𝐺(𝐵 + 𝐶)) = ((𝐴𝐺𝐵) + (𝐴𝐺𝐶)))
Theorem	sigarms 46307*	Signed area is additive (with respect to subtraction) by the second argument. (Contributed by Saveliy Skresanov, 19-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (𝐴𝐺(𝐵 − 𝐶)) = ((𝐴𝐺𝐵) − (𝐴𝐺𝐶)))
Theorem	sigarls 46308*	Signed area is linear by the second argument. (Contributed by Saveliy Skresanov, 19-Sep-2017.)
⊢ 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℝ) → (𝐴𝐺(𝐵 · 𝐶)) = ((𝐴𝐺𝐵) · 𝐶))
Theorem	sigarid 46309*	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 46310*	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 46311*	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 46312*	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 46313*	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 46314*	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 46315*	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 46316*	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 46317*	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 46318	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 46319*	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 46320*	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 46319 three times and then using cevathlem1 46318 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 46315. 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)))
21.41.2  Simple groups
Theorem	simpcntrab 46321	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))
21.42  Mathbox for Ender Ting
21.42.1  Increasing sequences and subsequences
Theorem	et-ltneverrefl 46322	Less-than class is never reflexive. (Contributed by Ender Ting, 22-Nov-2024.) Prefer to specify theorem domain and then apply ltnri 11353. (New usage is discouraged.)
⊢ ¬ 𝐴 < 𝐴
Theorem	et-equeucl 46323	Alternative proof that equality is left-Euclidean, using ax7 2011 directly instead of utility theorems; done for practice. (Contributed by Ender Ting, 21-Dec-2024.)
⊢ (𝑥 = 𝑧 → (𝑦 = 𝑧 → 𝑥 = 𝑦))
Theorem	et-sqrtnegnre 46324	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	natlocalincr 46325*	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 46326*	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)
Syntax	cupword 46327	Extend class notation to include the set of strictly increasing sequences.
class UpWord 𝑆
Definition	df-upword 46328*	Strictly increasing sequence is a sequence, adjacent elements of which increase. (Contributed by Ender Ting, 19-Nov-2024.)
⊢ UpWord 𝑆 = {𝑤 ∣ (𝑤 ∈ Word 𝑆 ∧ ∀𝑘 ∈ (0..^((♯‘𝑤) − 1))(𝑤‘𝑘) < (𝑤‘(𝑘 + 1)))}
Theorem	upwordnul 46329	Empty set is an increasing sequence for every range. (Contributed by Ender Ting, 19-Nov-2024.)
⊢ ∅ ∈ UpWord 𝑆
Theorem	upwordisword 46330	Any increasing sequence is a sequence. (Contributed by Ender Ting, 19-Nov-2024.)
⊢ (𝐴 ∈ UpWord 𝑆 → 𝐴 ∈ Word 𝑆)
Theorem	singoutnword 46331	Singleton with character out of range 𝑉 is not a word for that range. (Contributed by Ender Ting, 21-Nov-2024.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (¬ 𝐴 ∈ 𝑉 → ¬ ⟨“𝐴”⟩ ∈ Word 𝑉)
Theorem	singoutnupword 46332	Singleton with character out of range 𝑆 is not an increasing sequence for that range. (Contributed by Ender Ting, 22-Nov-2024.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (¬ 𝐴 ∈ 𝑆 → ¬ ⟨“𝐴”⟩ ∈ UpWord 𝑆)
Theorem	upwordsing 46333	Singleton is an increasing sequence for any compatible range. (Contributed by Ender Ting, 21-Nov-2024.)
⊢ 𝐴 ∈ 𝑆    ⇒   ⊢ ⟨“𝐴”⟩ ∈ UpWord 𝑆
Theorem	upwordsseti 46334	Strictly increasing sequences with a set given for range form a set. (Contributed by Ender Ting, 21-Nov-2024.)
⊢ 𝑆 ∈ V    ⇒   ⊢ UpWord 𝑆 ∈ V
Theorem	tworepnotupword 46335	Concatenation of identical singletons is never an increasing sequence. (Contributed by Ender Ting, 22-Nov-2024.)
⊢ 𝐴 ∈ V    ⇒   ⊢ ¬ (⟨“𝐴”⟩ ++ ⟨“𝐴”⟩) ∈ UpWord 𝑆
Theorem	upwrdfi 46336*	There is a finite number of strictly increasing sequences of a given length over finite alphabet. Trivially holds for invalid lengths where there're zero matching sequences. (Contributed by Ender Ting, 5-Jan-2024.)
⊢ (𝑆 ∈ Fin → {𝑎 ∈ UpWord 𝑆 ∣ (♯‘𝑎) = 𝑇} ∈ Fin)
21.43  Mathbox for Jarvin Udandy
Theorem	hirstL-ax3 46337	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 46338	Recover ax-3 8 from hirstL-ax3 46337. (Contributed by Jarvin Udandy, 3-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ ((¬ 𝜑 → ¬ 𝜓) → (𝜓 → 𝜑))
Theorem	aibandbiaiffaiffb 46339	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 46340	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 46341	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 46342	Given a is equivalent to ⊤, there exists a proof for a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
⊢ (𝜑 ↔ ⊤)    ⇒   ⊢ 𝜑
Theorem	aisfina 46343	Given a is equivalent to ⊥, there exists a proof for not a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
⊢ (𝜑 ↔ ⊥)    ⇒   ⊢ ¬ 𝜑
Theorem	bothtbothsame 46344	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 46345	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 46346	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 46347	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 46348	Given a is exclusive to b, there exists a proof for (not (a if-and-only-if b)); df-xor 1505 is a closed form of this. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜓)    ⇒   ⊢ ¬ (𝜑 ↔ 𝜓)
Theorem	aiffnbandciffatnotciffb 46349	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 46350	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 46351	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 46352	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 46353	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 46354	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 46355	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 46356	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 46357	If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.)
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))
Theorem	bisaiaisb 46358	Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ((𝜓 ↔ 𝜑) → (𝜑 ↔ 𝜓))
Theorem	atbiffatnnbalt 46359	If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))
Theorem	abnotbtaxb 46360	Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ 𝜑    &   ⊢ ¬ 𝜓    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	abnotataxb 46361	Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ¬ 𝜑    &   ⊢ 𝜓    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	conimpf 46362	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 46363	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 46364	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 46365	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 46366	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 46367	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 46368	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 46369	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 46370	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 46371	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 46372	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 46373	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 46374	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 46375	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 46376	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 46377	For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.)
⊢ (𝜑 ∧ 𝜓)    ⇒   ⊢ (𝜑 → 𝜓)
Theorem	onenotinotbothi 46378	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 46379	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 46380	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 46381	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 46382	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 46383	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 46384	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 46385	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 46386	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 46387	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 46388	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 46389	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 46390	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 46391	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 46392	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 46393	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 46394	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 46395	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 46396	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 46397	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 46398	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 46399	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 46400	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))