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Theorem List for Metamath Proof Explorer - 43301-43400   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremsmflimsup 43301* The superior limit of a sequence of sigma-measurable functions is sigma-measurable. Proposition 121F (d) of [Fremlin1] p. 39 . (Contributed by Glauco Siliprandi, 23-Oct-2021.)
𝑚𝐹    &   𝑥𝐹    &   (𝜑𝑀 ∈ ℤ)    &   𝑍 = (ℤ𝑀)    &   (𝜑𝑆 ∈ SAlg)    &   (𝜑𝐹:𝑍⟶(SMblFn‘𝑆))    &   𝐷 = {𝑥 𝑛𝑍 𝑚 ∈ (ℤ𝑛)dom (𝐹𝑚) ∣ (lim sup‘(𝑚𝑍 ↦ ((𝐹𝑚)‘𝑥))) ∈ ℝ}    &   𝐺 = (𝑥𝐷 ↦ (lim sup‘(𝑚𝑍 ↦ ((𝐹𝑚)‘𝑥))))       (𝜑𝐺 ∈ (SMblFn‘𝑆))

Theoremsmflimsupmpt 43302* The superior limit of a sequence of sigma-measurable functions is sigma-measurable. Proposition 121F (d) of [Fremlin1] p. 39 . 𝐴 can contain 𝑚 as a free variable, in other words it can be thought of as an indexed collection 𝐴(𝑚). 𝐵 can be thought of as a collection with two indices 𝐵(𝑚, 𝑥). (Contributed by Glauco Siliprandi, 23-Oct-2021.)
𝑚𝜑    &   𝑥𝜑    &   𝑛𝜑    &   (𝜑𝑀 ∈ ℤ)    &   𝑍 = (ℤ𝑀)    &   (𝜑𝑆 ∈ SAlg)    &   ((𝜑𝑚𝑍𝑥𝐴) → 𝐵𝑊)    &   ((𝜑𝑚𝑍) → (𝑥𝐴𝐵) ∈ (SMblFn‘𝑆))    &   𝐷 = {𝑥 𝑛𝑍 𝑚 ∈ (ℤ𝑛)𝐴 ∣ (lim sup‘(𝑚𝑍𝐵)) ∈ ℝ}    &   𝐺 = (𝑥𝐷 ↦ (lim sup‘(𝑚𝑍𝐵)))       (𝜑𝐺 ∈ (SMblFn‘𝑆))

Theoremsmfliminflem 43303* The inferior limit of a countable set of sigma-measurable functions is sigma-measurable. Proposition 121F (e) of [Fremlin1] p. 39 . (Contributed by Glauco Siliprandi, 2-Jan-2022.)
(𝜑𝑀 ∈ ℤ)    &   𝑍 = (ℤ𝑀)    &   (𝜑𝑆 ∈ SAlg)    &   (𝜑𝐹:𝑍⟶(SMblFn‘𝑆))    &   𝐷 = {𝑥 𝑛𝑍 𝑚 ∈ (ℤ𝑛)dom (𝐹𝑚) ∣ (lim inf‘(𝑚𝑍 ↦ ((𝐹𝑚)‘𝑥))) ∈ ℝ}    &   𝐺 = (𝑥𝐷 ↦ (lim inf‘(𝑚𝑍 ↦ ((𝐹𝑚)‘𝑥))))       (𝜑𝐺 ∈ (SMblFn‘𝑆))

Theoremsmfliminf 43304* The inferior limit of a countable set of sigma-measurable functions is sigma-measurable. Proposition 121F (e) of [Fremlin1] p. 39 . (Contributed by Glauco Siliprandi, 2-Jan-2022.)
𝑚𝐹    &   𝑥𝐹    &   (𝜑𝑀 ∈ ℤ)    &   𝑍 = (ℤ𝑀)    &   (𝜑𝑆 ∈ SAlg)    &   (𝜑𝐹:𝑍⟶(SMblFn‘𝑆))    &   𝐷 = {𝑥 𝑛𝑍 𝑚 ∈ (ℤ𝑛)dom (𝐹𝑚) ∣ (lim inf‘(𝑚𝑍 ↦ ((𝐹𝑚)‘𝑥))) ∈ ℝ}    &   𝐺 = (𝑥𝐷 ↦ (lim inf‘(𝑚𝑍 ↦ ((𝐹𝑚)‘𝑥))))       (𝜑𝐺 ∈ (SMblFn‘𝑆))

Theoremsmfliminfmpt 43305* The inferior limit of a countable set of sigma-measurable functions is sigma-measurable. Proposition 121F (e) of [Fremlin1] p. 39 . 𝐴 can contain 𝑚 as a free variable, in other words it can be thought of as an indexed collection 𝐴(𝑚). 𝐵 can be thought of as a collection with two indices 𝐵(𝑚, 𝑥). (Contributed by Glauco Siliprandi, 2-Jan-2022.)
𝑚𝜑    &   𝑥𝜑    &   𝑛𝜑    &   (𝜑𝑀 ∈ ℤ)    &   𝑍 = (ℤ𝑀)    &   (𝜑𝑆 ∈ SAlg)    &   ((𝜑𝑚𝑍𝑥𝐴) → 𝐵𝑉)    &   ((𝜑𝑚𝑍) → (𝑥𝐴𝐵) ∈ (SMblFn‘𝑆))    &   𝐷 = {𝑥 𝑛𝑍 𝑚 ∈ (ℤ𝑛)𝐴 ∣ (lim inf‘(𝑚𝑍𝐵)) ∈ ℝ}    &   𝐺 = (𝑥𝐷 ↦ (lim inf‘(𝑚𝑍𝐵)))       (𝜑𝐺 ∈ (SMblFn‘𝑆))

20.38  Mathbox for Saveliy Skresanov

20.38.1  Ceva's theorem

Theoremsigarval 43306* 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)))

Theoremsigarim 43307* 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)

Theoremsigarac 43308* 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))

Theoremsigaraf 43309* 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)))

Theoremsigarmf 43310* 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) → ((𝐴𝐶)𝐺𝐵) = ((𝐴𝐺𝐵) − (𝐶𝐺𝐵)))

Theoremsigaras 43311* 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)))

Theoremsigarms 43312* 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(𝐵𝐶)) = ((𝐴𝐺𝐵) − (𝐴𝐺𝐶)))

Theoremsigarls 43313* 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))

Theoremsigarid 43314* 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)

Theoremsigarexp 43315* 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) → ((𝐴𝐶)𝐺(𝐵𝐶)) = (((𝐴𝐺𝐵) − (𝐴𝐺𝐶)) − (𝐶𝐺𝐵)))

Theoremsigarperm 43316* Signed area (𝐴𝐶)𝐺(𝐵𝐶) 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) → ((𝐴𝐶)𝐺(𝐵𝐶)) = ((𝐵𝐴)𝐺(𝐶𝐴)))

Theoremsigardiv 43317* If signed area between vectors 𝐵𝐴 and 𝐶𝐴 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 → ((𝐵𝐴)𝐺(𝐶𝐴)) = 0)       (𝜑 → ((𝐵𝐴) / (𝐶𝐴)) ∈ ℝ)

Theoremsigarimcd 43318* 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)

Theoremsigariz 43319* 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)

Theoremsigarcol 43320* 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 → (((𝐴𝐶)𝐺(𝐵𝐶)) = 0 ↔ ∃𝑡 ∈ ℝ 𝐶 = (𝐵 + (𝑡 · (𝐴𝐵)))))

Theoremsharhght 43321* 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 ∧ ((𝐴𝐷)𝐺(𝐵𝐷)) = 0))       (𝜑 → (((𝐶𝐴)𝐺(𝐷𝐴)) · (𝐵𝐷)) = (((𝐶𝐵)𝐺(𝐷𝐵)) · (𝐴𝐷)))

Theoremsigaradd 43322* 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 ∧ ((𝐴𝐷)𝐺(𝐵𝐷)) = 0))       (𝜑 → (((𝐵𝐶)𝐺(𝐴𝐶)) − ((𝐷𝐶)𝐺(𝐴𝐶))) = ((𝐵𝐶)𝐺(𝐷𝐶)))

Theoremcevathlem1 43323 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))

Theoremcevathlem2 43324* 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))    &   (𝜑𝑂 ∈ ℂ)    &   (𝜑 → (((𝐴𝑂)𝐺(𝐷𝑂)) = 0 ∧ ((𝐵𝑂)𝐺(𝐸𝑂)) = 0 ∧ ((𝐶𝑂)𝐺(𝐹𝑂)) = 0))    &   (𝜑 → (((𝐴𝐹)𝐺(𝐵𝐹)) = 0 ∧ ((𝐵𝐷)𝐺(𝐶𝐷)) = 0 ∧ ((𝐶𝐸)𝐺(𝐴𝐸)) = 0))    &   (𝜑 → (((𝐴𝑂)𝐺(𝐵𝑂)) ≠ 0 ∧ ((𝐵𝑂)𝐺(𝐶𝑂)) ≠ 0 ∧ ((𝐶𝑂)𝐺(𝐴𝑂)) ≠ 0))       (𝜑 → (((𝐶𝑂)𝐺(𝐴𝑂)) · (𝐵𝐷)) = (((𝐴𝑂)𝐺(𝐵𝑂)) · (𝐷𝐶)))

Theoremcevath 43325* 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 43324 three times and then using cevathlem1 43323 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 43320. 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))    &   (𝜑𝑂 ∈ ℂ)    &   (𝜑 → (((𝐴𝑂)𝐺(𝐷𝑂)) = 0 ∧ ((𝐵𝑂)𝐺(𝐸𝑂)) = 0 ∧ ((𝐶𝑂)𝐺(𝐹𝑂)) = 0))    &   (𝜑 → (((𝐴𝐹)𝐺(𝐵𝐹)) = 0 ∧ ((𝐵𝐷)𝐺(𝐶𝐷)) = 0 ∧ ((𝐶𝐸)𝐺(𝐴𝐸)) = 0))    &   (𝜑 → (((𝐴𝑂)𝐺(𝐵𝑂)) ≠ 0 ∧ ((𝐵𝑂)𝐺(𝐶𝑂)) ≠ 0 ∧ ((𝐶𝑂)𝐺(𝐴𝑂)) ≠ 0))       (𝜑 → (((𝐴𝐹) · (𝐶𝐸)) · (𝐵𝐷)) = (((𝐹𝐵) · (𝐸𝐴)) · (𝐷𝐶)))

20.38.2  Simple groups

Theoremsimpcntrab 43326 The center of a simple group is trivial or the group is abelian. (Contributed by SS, 3-Jan-2024.)
𝐵 = (Base‘𝐺)    &    0 = (0g𝐺)    &   𝑍 = (Cntr‘𝐺)    &   (𝜑𝐺 ∈ SimpGrp)       (𝜑 → (𝑍 = { 0 } ∨ 𝐺 ∈ Abel))

20.39  Mathbox for Jarvin Udandy

TheoremhirstL-ax3 43327 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) → ((¬ 𝜑𝜓) → 𝜑))

Theoremax3h 43328 Recover ax-3 8 from hirstL-ax3 43327. (Contributed by Jarvin Udandy, 3-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
((¬ 𝜑 → ¬ 𝜓) → (𝜓𝜑))

Theoremaibandbiaiffaiffb 43329 A closed form showing (a implies b and b implies a) same-as (a same-as b). (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((𝜑𝜓) ∧ (𝜓𝜑)) ↔ (𝜑𝜓))

Theoremaibandbiaiaiffb 43330 A closed form showing (a implies b and b implies a) implies (a same-as b). (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((𝜑𝜓) ∧ (𝜓𝜑)) → (𝜑𝜓))

Theoremnotatnand 43331 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        ¬ (𝜑𝜓)

Theoremaistia 43332 Given a is equivalent to , there exists a proof for a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
(𝜑 ↔ ⊤)       𝜑

Theoremaisfina 43333 Given a is equivalent to , there exists a proof for not a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
(𝜑 ↔ ⊥)        ¬ 𝜑

Theorembothtbothsame 43334 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 ↔ ⊤)       (𝜑𝜓)

Theorembothfbothsame 43335 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 ↔ ⊥)       (𝜑𝜓)

Theoremaiffbbtat 43336 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.)
(𝜑𝜓)    &   (𝜓 ↔ ⊤)       (𝜑 ↔ ⊤)

Theoremaisbbisfaisf 43337 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.)
(𝜑𝜓)    &   (𝜓 ↔ ⊥)       (𝜑 ↔ ⊥)

Theoremaxorbtnotaiffb 43338 Given a is exclusive to b, there exists a proof for (not (a if-and-only-if b)); df-xor 1503 is a closed form of this. (Contributed by Jarvin Udandy, 7-Sep-2016.)
(𝜑𝜓)        ¬ (𝜑𝜓)

Theoremaiffnbandciffatnotciffb 43339 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)    &   (𝜒𝜑)        ¬ (𝜒𝜓)

Theoremaxorbciffatcxorb 43340 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.)
(𝜑𝜓)    &   (𝜒𝜑)       (𝜒𝜓)

Theoremaibnbna 43341 Given a implies b, (not b), there exists a proof for (not a). (Contributed by Jarvin Udandy, 1-Sep-2016.)
(𝜑𝜓)    &    ¬ 𝜓        ¬ 𝜑

Theoremaibnbaif 43342 Given a implies b, not b, there exists a proof for a is F. (Contributed by Jarvin Udandy, 1-Sep-2016.)
(𝜑𝜓)    &    ¬ 𝜓       (𝜑 ↔ ⊥)

Theoremaiffbtbat 43343 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.)
(𝜑𝜓)    &   (⊤ ↔ 𝜓)       (𝜑 ↔ ⊤)

Theoremastbstanbst 43344 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 ↔ ⊤)       ((𝜑𝜓) ↔ ⊤)

Theoremaistbistaandb 43345 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 ↔ ⊤)       (𝜑𝜓)

Theoremaisbnaxb 43346 Given a is equivalent to b, there exists a proof for (not (a xor b)). (Contributed by Jarvin Udandy, 28-Aug-2016.)
(𝜑𝜓)        ¬ (𝜑𝜓)

Theorematbiffatnnb 43347 If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.)
((𝜑𝜓) → (𝜑 → ¬ ¬ 𝜓))

Theorembisaiaisb 43348 Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.)
((𝜓𝜑) → (𝜑𝜓))

Theorematbiffatnnbalt 43349 If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.)
((𝜑𝜓) → (𝜑 → ¬ ¬ 𝜓))

Theoremabnotbtaxb 43350 Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
𝜑    &    ¬ 𝜓       (𝜑𝜓)

Theoremabnotataxb 43351 Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
¬ 𝜑    &   𝜓       (𝜑𝜓)

Theoremconimpf 43352 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    &   (𝜑𝜓)       (𝜑 ↔ ⊥)

Theoremconimpfalt 43353 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    &   (𝜑𝜓)       (𝜑 ↔ ⊥)

Theoremaistbisfiaxb 43354 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 ↔ ⊥)       (𝜑𝜓)

Theoremaisfbistiaxb 43355 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 ↔ ⊤)       (𝜑𝜓)

Theoremaifftbifffaibif 43356 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 ↔ ⊥)       ((𝜑𝜓) ↔ ⊥)

Theoremaifftbifffaibifff 43357 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 ↔ ⊥)       ((𝜑𝜓) ↔ ⊥)

Theorematnaiana 43358 Given a, it is not the case a implies a self contradiction. (Contributed by Jarvin Udandy, 7-Sep-2020.)
𝜑        ¬ (𝜑 → (𝜑 ∧ ¬ 𝜑))

Theoremainaiaandna 43359 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)))

Theoremabcdta 43360 Given (((a and b) and c) and d), there exists a proof for a. (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((𝜑𝜓) ∧ 𝜒) ∧ 𝜃)       𝜑

Theoremabcdtb 43361 Given (((a and b) and c) and d), there exists a proof for b. (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((𝜑𝜓) ∧ 𝜒) ∧ 𝜃)       𝜓

Theoremabcdtc 43362 Given (((a and b) and c) and d), there exists a proof for c. (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((𝜑𝜓) ∧ 𝜒) ∧ 𝜃)       𝜒

Theoremabcdtd 43363 Given (((a and b) and c) and d), there exists a proof for d. (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((𝜑𝜓) ∧ 𝜒) ∧ 𝜃)       𝜃

Theoremabciffcbatnabciffncba 43364 Operands in a biconditional expression converted negated. Additionally biconditional converted to show antecedent implies sequent. Closed form. (Contributed by Jarvin Udandy, 7-Sep-2020.)
(¬ ((𝜑𝜓) ∧ 𝜒) → ¬ ((𝜒𝜓) ∧ 𝜑))

Theoremabciffcbatnabciffncbai 43365 Operands in a biconditional expression converted negated. Additionally biconditional converted to show antecedent implies sequent. (Contributed by Jarvin Udandy, 7-Sep-2020.)
(((𝜑𝜓) ∧ 𝜒) ↔ ((𝜒𝜓) ∧ 𝜑))       (¬ ((𝜑𝜓) ∧ 𝜒) → ¬ ((𝜒𝜓) ∧ 𝜑))

Theoremnabctnabc 43366 not ( a -> ( b /\ c ) ) we can show: not a implies ( b /\ c ). (Contributed by Jarvin Udandy, 7-Sep-2020.)
¬ (𝜑 → (𝜓𝜒))       𝜑 → (𝜓𝜒))

Theoremjabtaib 43367 For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.)
(𝜑𝜓)       (𝜑𝜓)

Theoremonenotinotbothi 43368 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.)
¬ (𝜑𝜓)        ¬ ((𝜑𝜓) ∧ (𝜒𝜃))

Theoremtwonotinotbothi 43369 From these two negated implications it is not the case their nonnegated forms are both true. (Contributed by Jarvin Udandy, 11-Sep-2020.)
¬ (𝜑𝜓)    &    ¬ (𝜒𝜃)        ¬ ((𝜑𝜓) ∧ (𝜒𝜃))

Theoremclifte 43370 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) ∨ (𝜓𝜒)))

Theoremcliftet 43371 show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
(𝜑𝜒)    &   𝜃       (𝜃 ↔ ((𝜑𝜒) ∨ (𝜓 ∧ ¬ 𝜒)))

Theoremclifteta 43372 show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
((𝜑 ∧ ¬ 𝜒) ∨ (𝜓𝜒))    &   𝜃       (𝜃 ↔ ((𝜑 ∧ ¬ 𝜒) ∨ (𝜓𝜒)))

Theoremcliftetb 43373 show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
((𝜑𝜒) ∨ (𝜓 ∧ ¬ 𝜒))    &   𝜃       (𝜃 ↔ ((𝜑𝜒) ∨ (𝜓 ∧ ¬ 𝜒)))

Theoremconfun 43374 Given the hypotheses there exists a proof for (c implies ( d iff a ) ). (Contributed by Jarvin Udandy, 6-Sep-2020.)
𝜑    &   (𝜒𝜓)    &   (𝜒𝜃)    &   (𝜑 → (𝜑𝜓))       (𝜒 → (𝜃𝜑))

Theoremconfun2 43375 Confun simplified to two propositions. (Contributed by Jarvin Udandy, 6-Sep-2020.)
(𝜓𝜑)    &   (𝜓 → ¬ (𝜓 → (𝜓 ∧ ¬ 𝜓)))    &   ((𝜓𝜑) → ((𝜓𝜑) → 𝜑))       (𝜓 → (¬ (𝜓 → (𝜓 ∧ ¬ 𝜓)) ↔ (𝜓𝜑)))

Theoremconfun3 43376 Confun's more complex form where both a,d have been "defined". (Contributed by Jarvin Udandy, 6-Sep-2020.)
(𝜑 ↔ (𝜒𝜓))    &   (𝜃 ↔ ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   (𝜒𝜓)    &   (𝜒 → ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   ((𝜒𝜓) → ((𝜒𝜓) → 𝜓))       (𝜒 → (¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)) ↔ (𝜒𝜓)))

Theoremconfun4 43377 An attempt at derivative. Resisted simplest path to a proof. (Contributed by Jarvin Udandy, 6-Sep-2020.)
𝜑    &   ((𝜑𝜓) → 𝜓)    &   (𝜓 → (𝜑𝜒))    &   ((𝜒𝜃) → ((𝜑𝜃) ↔ 𝜓))    &   (𝜏 ↔ (𝜒𝜃))    &   (𝜂 ↔ ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   𝜓    &   (𝜒𝜃)       (𝜒 → (𝜓𝜏))

Theoremconfun5 43378 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    &   ((𝜑𝜓) → 𝜓)    &   (𝜓 → (𝜑𝜒))    &   ((𝜒𝜃) → ((𝜑𝜃) ↔ 𝜓))    &   (𝜏 ↔ (𝜒𝜃))    &   (𝜂 ↔ ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   𝜓    &   (𝜒𝜃)       (𝜒 → (𝜂𝜏))

Theoremplcofph 43379 Given, a,b and a "definition" for c, c is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
(𝜒 ↔ ((((𝜑𝜓) ↔ 𝜑) → (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))) ∧ (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))))    &   𝜑    &   𝜓       𝜒

Theorempldofph 43380 Given, a,b c, d, "definition" for e, e is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
(𝜏 ↔ ((𝜒𝜃) ∧ (𝜑𝜒) ∧ ((𝜑𝜓) → (𝜓𝜃))))    &   𝜑    &   𝜓    &   𝜒    &   𝜃       𝜏

Theoremplvcofph 43381 Given, a,b,d, and "definitions" for c, e, f: f is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
(𝜒 ↔ ((((𝜑𝜓) ↔ 𝜑) → (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))) ∧ (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))))    &   (𝜏 ↔ ((𝜒𝜃) ∧ (𝜑𝜒) ∧ ((𝜑𝜓) → (𝜓𝜃))))    &   (𝜂 ↔ (𝜒𝜏))    &   𝜑    &   𝜓    &   𝜃       𝜂

Theoremplvcofphax 43382 Given, a,b,d, and "definitions" for c, e, f, g: g is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
(𝜒 ↔ ((((𝜑𝜓) ↔ 𝜑) → (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))) ∧ (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))))    &   (𝜏 ↔ ((𝜒𝜃) ∧ (𝜑𝜒) ∧ ((𝜑𝜓) → (𝜓𝜃))))    &   (𝜂 ↔ (𝜒𝜏))    &   𝜑    &   𝜓    &   𝜃    &   (𝜁 ↔ ¬ (𝜓 ∧ ¬ 𝜏))       𝜁

Theoremplvofpos 43383 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 ↔ (¬ 𝜑𝜓))    &   (𝜏 ↔ (𝜑 ∧ ¬ 𝜓))    &   (𝜂 ↔ (𝜑𝜓))    &   (𝜁 ↔ (((((¬ ((𝜇𝜒) ∧ (𝜇𝜃)) ∧ ¬ ((𝜇𝜒) ∧ (𝜇𝜏))) ∧ ¬ ((𝜇𝜒) ∧ (𝜒𝜂))) ∧ ¬ ((𝜇𝜃) ∧ (𝜇𝜏))) ∧ ¬ ((𝜇𝜃) ∧ (𝜇𝜂))) ∧ ¬ ((𝜇𝜏) ∧ (𝜇𝜂))))    &   (𝜎 ↔ (((𝜇𝜒) ∨ (𝜇𝜃)) ∨ ((𝜇𝜏) ∨ (𝜇𝜂))))    &   (𝜌 ↔ (𝜁𝜎))    &   𝜁    &   𝜎       𝜌

Theoremmdandyv0 43384 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 ↔ ⊥)       ((((𝜒𝜑) ∧ (𝜃𝜑)) ∧ (𝜏𝜑)) ∧ (𝜂𝜑))

Theoremmdandyv1 43385 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 ↔ ⊥)       ((((𝜒𝜓) ∧ (𝜃𝜑)) ∧ (𝜏𝜑)) ∧ (𝜂𝜑))

Theoremmdandyv2 43386 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 ↔ ⊥)       ((((𝜒𝜑) ∧ (𝜃𝜓)) ∧ (𝜏𝜑)) ∧ (𝜂𝜑))

Theoremmdandyv3 43387 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 ↔ ⊥)       ((((𝜒𝜓) ∧ (𝜃𝜓)) ∧ (𝜏𝜑)) ∧ (𝜂𝜑))

Theoremmdandyv4 43388 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 ↔ ⊥)       ((((𝜒𝜑) ∧ (𝜃𝜑)) ∧ (𝜏𝜓)) ∧ (𝜂𝜑))

Theoremmdandyv5 43389 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 ↔ ⊥)       ((((𝜒𝜓) ∧ (𝜃𝜑)) ∧ (𝜏𝜓)) ∧ (𝜂𝜑))

Theoremmdandyv6 43390 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 ↔ ⊥)       ((((𝜒𝜑) ∧ (𝜃𝜓)) ∧ (𝜏𝜓)) ∧ (𝜂𝜑))

Theoremmdandyv7 43391 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 ↔ ⊥)       ((((𝜒𝜓) ∧ (𝜃𝜓)) ∧ (𝜏𝜓)) ∧ (𝜂𝜑))

Theoremmdandyv8 43392 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 ↔ ⊤)       ((((𝜒𝜑) ∧ (𝜃𝜑)) ∧ (𝜏𝜑)) ∧ (𝜂𝜓))

Theoremmdandyv9 43393 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 ↔ ⊤)       ((((𝜒𝜓) ∧ (𝜃𝜑)) ∧ (𝜏𝜑)) ∧ (𝜂𝜓))

Theoremmdandyv10 43394 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 ↔ ⊤)       ((((𝜒𝜑) ∧ (𝜃𝜓)) ∧ (𝜏𝜑)) ∧ (𝜂𝜓))

Theoremmdandyv11 43395 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 ↔ ⊤)       ((((𝜒𝜓) ∧ (𝜃𝜓)) ∧ (𝜏𝜑)) ∧ (𝜂𝜓))

Theoremmdandyv12 43396 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 ↔ ⊤)       ((((𝜒𝜑) ∧ (𝜃𝜑)) ∧ (𝜏𝜓)) ∧ (𝜂𝜓))

Theoremmdandyv13 43397 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 ↔ ⊤)       ((((𝜒𝜓) ∧ (𝜃𝜑)) ∧ (𝜏𝜓)) ∧ (𝜂𝜓))

Theoremmdandyv14 43398 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 ↔ ⊤)       ((((𝜒𝜑) ∧ (𝜃𝜓)) ∧ (𝜏𝜓)) ∧ (𝜂𝜓))

Theoremmdandyv15 43399 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 ↔ ⊤)       ((((𝜒𝜓) ∧ (𝜃𝜓)) ∧ (𝜏𝜓)) ∧ (𝜂𝜓))

Theoremmdandyvr0 43400 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜑)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜁) ∧ (𝜃𝜁)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))

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