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Theorem List for Metamath Proof Explorer - 39701-39800   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremsigariz 39701* 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 39702* 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 39703* 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 39704* 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 39705 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 39706* 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 39707* 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 39706 three times and then using cevathlem1 39705 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 39702. 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)))

21.33  Mathbox for Jarvin Udandy

TheoremhirstL-ax3 39708 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 39709 Recovery of ax-3 8 from hirstL-ax3 39708. (Contributed by Jarvin Udandy, 3-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
((¬ 𝜑 → ¬ 𝜓) → (𝜓𝜑))

Theoremaibandbiaiffaiffb 39710 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 39711 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 39712 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 39713 Given a is equivalent to , there exists a proof for a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
(𝜑 ↔ ⊤)       𝜑

Theoremaisfina 39714 Given a is equivalent to , there exists a proof for not a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
(𝜑 ↔ ⊥)        ¬ 𝜑

Theorembothtbothsame 39715 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 39716 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 39717 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 39718 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 39719 Given a is exclusive to b, there exists a proof for (not (a if-and-only-if b)); df-xor 1457 is a closed form of this. (Contributed by Jarvin Udandy, 7-Sep-2016.)
(𝜑𝜓)        ¬ (𝜑𝜓)

Theoremaiffnbandciffatnotciffb 39720 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 39721 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 39722 Given a implies b, (not b), there exists a proof for (not a). (Contributed by Jarvin Udandy, 1-Sep-2016.)
(𝜑𝜓)    &    ¬ 𝜓        ¬ 𝜑

Theoremaibnbaif 39723 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 39724 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 39725 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 39726 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 39727 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 39728 If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.)
((𝜑𝜓) → (𝜑 → ¬ ¬ 𝜓))

Theorembisaiaisb 39729 Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.)
((𝜓𝜑) → (𝜑𝜓))

Theorematbiffatnnbalt 39730 If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.)
((𝜑𝜓) → (𝜑 → ¬ ¬ 𝜓))

Theoremabnotbtaxb 39731 Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
𝜑    &    ¬ 𝜓       (𝜑𝜓)

Theoremabnotataxb 39732 Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
¬ 𝜑    &   𝜓       (𝜑𝜓)

Theoremconimpf 39733 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 39734 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 39735 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 39736 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 39737 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 39738 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 39739 Given a, it is not the case a implies a self contradiction. (Contributed by Jarvin Udandy, 7-Sep-2020.)
𝜑        ¬ (𝜑 → (𝜑 ∧ ¬ 𝜑))

Theoremainaiaandna 39740 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 39741 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 39742 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 39743 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 39744 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 39745 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 39746 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 39747 not ( a -> ( b /\ c ) ) we can show: not a implies ( b /\ c ). (Contributed by Jarvin Udandy, 7-Sep-2020.)
¬ (𝜑 → (𝜓𝜒))       𝜑 → (𝜓𝜒))

Theoremjabtaib 39748 For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.)
(𝜑𝜓)       (𝜑𝜓)

Theoremonenotinotbothi 39749 From one negated implication it is not the case its non negated form and a random others are both true. (Contributed by Jarvin Udandy, 11-Sep-2020.)
¬ (𝜑𝜓)        ¬ ((𝜑𝜓) ∧ (𝜒𝜃))

Theoremtwonotinotbothi 39750 From these two negated implications it is not the case their non negated forms are both true. (Contributed by Jarvin Udandy, 11-Sep-2020.)
¬ (𝜑𝜓)    &    ¬ (𝜒𝜃)        ¬ ((𝜑𝜓) ∧ (𝜒𝜃))

Theoremclifte 39751 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 39752 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 39753 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 39754 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 39755 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 39756 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 39757 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 39758 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 39759 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 39760 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 39761 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 39762 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 39763 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 39764 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 39765 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 39766 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 39767 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 39768 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 39769 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 39770 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 39771 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 39772 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 39773 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 39774 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 39775 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 39776 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 39777 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 39778 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 39779 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 39780 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 39781 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))

Theoremmdandyvr1 39782 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)    &   (𝜒𝜓)    &   (𝜃𝜑)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜎) ∧ (𝜃𝜁)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))

Theoremmdandyvr2 39783 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)    &   (𝜃𝜓)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜁) ∧ (𝜃𝜎)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))

Theoremmdandyvr3 39784 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)    &   (𝜒𝜓)    &   (𝜃𝜓)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜎) ∧ (𝜃𝜎)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))

Theoremmdandyvr4 39785 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)    &   (𝜏𝜓)    &   (𝜂𝜑)       ((((𝜒𝜁) ∧ (𝜃𝜁)) ∧ (𝜏𝜎)) ∧ (𝜂𝜁))

Theoremmdandyvr5 39786 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)    &   (𝜒𝜓)    &   (𝜃𝜑)    &   (𝜏𝜓)    &   (𝜂𝜑)       ((((𝜒𝜎) ∧ (𝜃𝜁)) ∧ (𝜏𝜎)) ∧ (𝜂𝜁))

Theoremmdandyvr6 39787 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)    &   (𝜃𝜓)    &   (𝜏𝜓)    &   (𝜂𝜑)       ((((𝜒𝜁) ∧ (𝜃𝜎)) ∧ (𝜏𝜎)) ∧ (𝜂𝜁))

Theoremmdandyvr7 39788 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)    &   (𝜒𝜓)    &   (𝜃𝜓)    &   (𝜏𝜓)    &   (𝜂𝜑)       ((((𝜒𝜎) ∧ (𝜃𝜎)) ∧ (𝜏𝜎)) ∧ (𝜂𝜁))

Theoremmdandyvr8 39789 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)    &   (𝜂𝜓)       ((((𝜒𝜁) ∧ (𝜃𝜁)) ∧ (𝜏𝜁)) ∧ (𝜂𝜎))

Theoremmdandyvr9 39790 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)    &   (𝜒𝜓)    &   (𝜃𝜑)    &   (𝜏𝜑)    &   (𝜂𝜓)       ((((𝜒𝜎) ∧ (𝜃𝜁)) ∧ (𝜏𝜁)) ∧ (𝜂𝜎))

Theoremmdandyvr10 39791 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)    &   (𝜃𝜓)    &   (𝜏𝜑)    &   (𝜂𝜓)       ((((𝜒𝜁) ∧ (𝜃𝜎)) ∧ (𝜏𝜁)) ∧ (𝜂𝜎))

Theoremmdandyvr11 39792 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)    &   (𝜒𝜓)    &   (𝜃𝜓)    &   (𝜏𝜑)    &   (𝜂𝜓)       ((((𝜒𝜎) ∧ (𝜃𝜎)) ∧ (𝜏𝜁)) ∧ (𝜂𝜎))

Theoremmdandyvr12 39793 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)    &   (𝜏𝜓)    &   (𝜂𝜓)       ((((𝜒𝜁) ∧ (𝜃𝜁)) ∧ (𝜏𝜎)) ∧ (𝜂𝜎))

Theoremmdandyvr13 39794 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)    &   (𝜒𝜓)    &   (𝜃𝜑)    &   (𝜏𝜓)    &   (𝜂𝜓)       ((((𝜒𝜎) ∧ (𝜃𝜁)) ∧ (𝜏𝜎)) ∧ (𝜂𝜎))

Theoremmdandyvr14 39795 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)    &   (𝜃𝜓)    &   (𝜏𝜓)    &   (𝜂𝜓)       ((((𝜒𝜁) ∧ (𝜃𝜎)) ∧ (𝜏𝜎)) ∧ (𝜂𝜎))

Theoremmdandyvr15 39796 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)    &   (𝜒𝜓)    &   (𝜃𝜓)    &   (𝜏𝜓)    &   (𝜂𝜓)       ((((𝜒𝜎) ∧ (𝜃𝜎)) ∧ (𝜏𝜎)) ∧ (𝜂𝜎))

Theoremmdandyvrx0 39797 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜑)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜁) ∧ (𝜃𝜁)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))

Theoremmdandyvrx1 39798 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜑)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜎) ∧ (𝜃𝜁)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))

Theoremmdandyvrx2 39799 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜓)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜁) ∧ (𝜃𝜎)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))

Theoremmdandyvrx3 39800 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜓)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜎) ∧ (𝜃𝜎)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))

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