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Type | Label | Description |
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Statement | ||
Theorem | smfdivdmmbl2 46201 | If a functions and a sigma-measurable function have domains in the sigma-algebra, the domain of the division of the two functions is in the sigma-algebra. This is the third statement of Proposition 121H of [Fremlin1] p. 39 . Note: While the theorem in the book assumes both functions are sigma-measurable, this assumption is unnecessary for the part concerning their division, for the function at the numerator. It is required only for the function at the denominator. (Contributed by Glauco Siliprandi, 5-Jan-2025.) |
⊢ Ⅎ𝑥𝜑 & ⊢ Ⅎ𝑥𝐹 & ⊢ Ⅎ𝑥𝐺 & ⊢ (𝜑 → 𝑆 ∈ SAlg) & ⊢ (𝜑 → 𝐹:𝐴⟶𝑉) & ⊢ (𝜑 → 𝐺 ∈ (SMblFn‘𝑆)) & ⊢ (𝜑 → 𝐴 ∈ 𝑆) & ⊢ (𝜑 → dom 𝐺 ∈ 𝑆) & ⊢ 𝐷 = {𝑥 ∈ dom 𝐺 ∣ (𝐺‘𝑥) ≠ 0} & ⊢ 𝐻 = (𝑥 ∈ (dom 𝐹 ∩ 𝐷) ↦ ((𝐹‘𝑥) / (𝐺‘𝑥))) ⇒ ⊢ (𝜑 → dom 𝐻 ∈ 𝑆) | ||
Theorem | fsupdm 46202* | The domain of the sup function is defined in Proposition 121F (b) of [Fremlin1], p. 38. Note that this definition of the sup function is quite general, as it does not require the original functions to be sigma-measurable, and it could be applied to uncountable sets of functions. The equality proved here is part of the proof of the fourth statement of Proposition 121H in [Fremlin1], p. 39. (Contributed by Glauco Siliprandi, 24-Jan-2025.) |
⊢ Ⅎ𝑛𝜑 & ⊢ Ⅎ𝑥𝜑 & ⊢ Ⅎ𝑚𝜑 & ⊢ Ⅎ𝑥𝐹 & ⊢ ((𝜑 ∧ 𝑛 ∈ 𝑍) → (𝐹‘𝑛):dom (𝐹‘𝑛)⟶ℝ*) & ⊢ 𝐷 = {𝑥 ∈ ∩ 𝑛 ∈ 𝑍 dom (𝐹‘𝑛) ∣ ∃𝑦 ∈ ℝ ∀𝑛 ∈ 𝑍 ((𝐹‘𝑛)‘𝑥) ≤ 𝑦} & ⊢ 𝐻 = (𝑛 ∈ 𝑍 ↦ (𝑚 ∈ ℕ ↦ {𝑥 ∈ dom (𝐹‘𝑛) ∣ ((𝐹‘𝑛)‘𝑥) < 𝑚})) ⇒ ⊢ (𝜑 → 𝐷 = ∪ 𝑚 ∈ ℕ ∩ 𝑛 ∈ 𝑍 ((𝐻‘𝑛)‘𝑚)) | ||
Theorem | fsupdm2 46203* | The domain of the sup function is defined in Proposition 121F (b) of [Fremlin1], p. 38. Note that this definition of the sup function is quite general, as it does not require the original functions to be sigma-measurable, and it could be applied to uncountable sets of functions. The equality proved here is part of the proof of the fourth statement of Proposition 121H in [Fremlin1], p. 39. (Contributed by Glauco Siliprandi, 24-Jan-2025.) |
⊢ Ⅎ𝑛𝜑 & ⊢ Ⅎ𝑥𝜑 & ⊢ Ⅎ𝑚𝜑 & ⊢ Ⅎ𝑥𝐹 & ⊢ ((𝜑 ∧ 𝑛 ∈ 𝑍) → (𝐹‘𝑛):dom (𝐹‘𝑛)⟶ℝ*) & ⊢ 𝐷 = {𝑥 ∈ ∩ 𝑛 ∈ 𝑍 dom (𝐹‘𝑛) ∣ ∃𝑦 ∈ ℝ ∀𝑛 ∈ 𝑍 ((𝐹‘𝑛)‘𝑥) ≤ 𝑦} & ⊢ 𝐺 = (𝑥 ∈ 𝐷 ↦ sup(ran (𝑛 ∈ 𝑍 ↦ ((𝐹‘𝑛)‘𝑥)), ℝ, < )) & ⊢ 𝐻 = (𝑛 ∈ 𝑍 ↦ (𝑚 ∈ ℕ ↦ {𝑥 ∈ dom (𝐹‘𝑛) ∣ ((𝐹‘𝑛)‘𝑥) < 𝑚})) ⇒ ⊢ (𝜑 → dom 𝐺 = ∪ 𝑚 ∈ ℕ ∩ 𝑛 ∈ 𝑍 ((𝐻‘𝑛)‘𝑚)) | ||
Theorem | smfsupdmmbllem 46204* | If a countable set of sigma-measurable functions have domains in the sigma-algebra, then their supremum function has the domain in the sigma-algebra. This is the fourth statement of Proposition 121H of [Fremlin1] p. 39 . (Contributed by Glauco Siliprandi, 24-Jan-2025.) |
⊢ Ⅎ𝑛𝜑 & ⊢ Ⅎ𝑥𝜑 & ⊢ Ⅎ𝑚𝜑 & ⊢ Ⅎ𝑥𝐹 & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ 𝑍 = (ℤ≥‘𝑀) & ⊢ (𝜑 → 𝑆 ∈ SAlg) & ⊢ (𝜑 → 𝐹:𝑍⟶(SMblFn‘𝑆)) & ⊢ ((𝜑 ∧ 𝑛 ∈ 𝑍) → dom (𝐹‘𝑛) ∈ 𝑆) & ⊢ 𝐷 = {𝑥 ∈ ∩ 𝑛 ∈ 𝑍 dom (𝐹‘𝑛) ∣ ∃𝑦 ∈ ℝ ∀𝑛 ∈ 𝑍 ((𝐹‘𝑛)‘𝑥) ≤ 𝑦} & ⊢ 𝐻 = (𝑛 ∈ 𝑍 ↦ (𝑚 ∈ ℕ ↦ {𝑥 ∈ dom (𝐹‘𝑛) ∣ ((𝐹‘𝑛)‘𝑥) < 𝑚})) & ⊢ 𝐺 = (𝑥 ∈ 𝐷 ↦ sup(ran (𝑛 ∈ 𝑍 ↦ ((𝐹‘𝑛)‘𝑥)), ℝ, < )) ⇒ ⊢ (𝜑 → dom 𝐺 ∈ 𝑆) | ||
Theorem | smfsupdmmbl 46205* | If a countable set of sigma-measurable functions have domains in the sigma-algebra, then their supremum function has the domain in the sigma-algebra. This is the fourth statement of Proposition 121H of [Fremlin1] p. 39 . (Contributed by Glauco Siliprandi, 24-Jan-2025.) |
⊢ Ⅎ𝑛𝜑 & ⊢ Ⅎ𝑥𝜑 & ⊢ Ⅎ𝑥𝐹 & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ 𝑍 = (ℤ≥‘𝑀) & ⊢ (𝜑 → 𝑆 ∈ SAlg) & ⊢ (𝜑 → 𝐹:𝑍⟶(SMblFn‘𝑆)) & ⊢ ((𝜑 ∧ 𝑛 ∈ 𝑍) → dom (𝐹‘𝑛) ∈ 𝑆) & ⊢ 𝐷 = {𝑥 ∈ ∩ 𝑛 ∈ 𝑍 dom (𝐹‘𝑛) ∣ ∃𝑦 ∈ ℝ ∀𝑛 ∈ 𝑍 ((𝐹‘𝑛)‘𝑥) ≤ 𝑦} & ⊢ 𝐺 = (𝑥 ∈ 𝐷 ↦ sup(ran (𝑛 ∈ 𝑍 ↦ ((𝐹‘𝑛)‘𝑥)), ℝ, < )) ⇒ ⊢ (𝜑 → dom 𝐺 ∈ 𝑆) | ||
Theorem | finfdm 46206* | The domain of the inf function is defined in Proposition 121F (c) of [Fremlin1], p. 39. See smfinf 46178. Note that this definition of the inf function is quite general, as it does not require the original functions to be sigma-measurable, and it could be applied to uncountable sets of functions. The equality proved here is part of the proof of the fifth statement of Proposition 121H in [Fremlin1], p. 39. (Contributed by Glauco Siliprandi, 1-Feb-2025.) |
⊢ Ⅎ𝑛𝜑 & ⊢ Ⅎ𝑥𝜑 & ⊢ Ⅎ𝑚𝜑 & ⊢ Ⅎ𝑥𝐹 & ⊢ ((𝜑 ∧ 𝑛 ∈ 𝑍) → (𝐹‘𝑛):dom (𝐹‘𝑛)⟶ℝ*) & ⊢ 𝐷 = {𝑥 ∈ ∩ 𝑛 ∈ 𝑍 dom (𝐹‘𝑛) ∣ ∃𝑦 ∈ ℝ ∀𝑛 ∈ 𝑍 𝑦 ≤ ((𝐹‘𝑛)‘𝑥)} & ⊢ 𝐻 = (𝑛 ∈ 𝑍 ↦ (𝑚 ∈ ℕ ↦ {𝑥 ∈ dom (𝐹‘𝑛) ∣ -𝑚 < ((𝐹‘𝑛)‘𝑥)})) ⇒ ⊢ (𝜑 → 𝐷 = ∪ 𝑚 ∈ ℕ ∩ 𝑛 ∈ 𝑍 ((𝐻‘𝑛)‘𝑚)) | ||
Theorem | finfdm2 46207* | The domain of the inf function is defined in Proposition 121F (c) of [Fremlin1], p. 39. See smfinf 46178. Note that this definition of the inf function is quite general, as it does not require the original functions to be sigma-measurable, and it could be applied to uncountable sets of functions. The equality proved here is part of the proof of the fifth statement of Proposition 121H in [Fremlin1], p. 39. (Contributed by Glauco Siliprandi, 1-Feb-2025.) |
⊢ Ⅎ𝑛𝜑 & ⊢ Ⅎ𝑥𝜑 & ⊢ Ⅎ𝑚𝜑 & ⊢ Ⅎ𝑥𝐹 & ⊢ ((𝜑 ∧ 𝑛 ∈ 𝑍) → (𝐹‘𝑛):dom (𝐹‘𝑛)⟶ℝ*) & ⊢ 𝐷 = {𝑥 ∈ ∩ 𝑛 ∈ 𝑍 dom (𝐹‘𝑛) ∣ ∃𝑦 ∈ ℝ ∀𝑛 ∈ 𝑍 𝑦 ≤ ((𝐹‘𝑛)‘𝑥)} & ⊢ 𝐺 = (𝑥 ∈ 𝐷 ↦ inf(ran (𝑛 ∈ 𝑍 ↦ ((𝐹‘𝑛)‘𝑥)), ℝ, < )) & ⊢ 𝐻 = (𝑛 ∈ 𝑍 ↦ (𝑚 ∈ ℕ ↦ {𝑥 ∈ dom (𝐹‘𝑛) ∣ -𝑚 < ((𝐹‘𝑛)‘𝑥)})) ⇒ ⊢ (𝜑 → dom 𝐺 = ∪ 𝑚 ∈ ℕ ∩ 𝑛 ∈ 𝑍 ((𝐻‘𝑛)‘𝑚)) | ||
Theorem | smfinfdmmbllem 46208* | If a countable set of sigma-measurable functions have domains in the sigma-algebra, then their infimum function has the domain in the sigma-algebra. This is the fifth statement of Proposition 121H of [Fremlin1] p. 39 . (Contributed by Glauco Siliprandi, 1-Feb-2025.) |
⊢ Ⅎ𝑛𝜑 & ⊢ Ⅎ𝑥𝜑 & ⊢ Ⅎ𝑚𝜑 & ⊢ Ⅎ𝑥𝐹 & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ 𝑍 = (ℤ≥‘𝑀) & ⊢ (𝜑 → 𝑆 ∈ SAlg) & ⊢ (𝜑 → 𝐹:𝑍⟶(SMblFn‘𝑆)) & ⊢ ((𝜑 ∧ 𝑛 ∈ 𝑍) → dom (𝐹‘𝑛) ∈ 𝑆) & ⊢ 𝐷 = {𝑥 ∈ ∩ 𝑛 ∈ 𝑍 dom (𝐹‘𝑛) ∣ ∃𝑦 ∈ ℝ ∀𝑛 ∈ 𝑍 𝑦 ≤ ((𝐹‘𝑛)‘𝑥)} & ⊢ 𝐺 = (𝑥 ∈ 𝐷 ↦ inf(ran (𝑛 ∈ 𝑍 ↦ ((𝐹‘𝑛)‘𝑥)), ℝ, < )) & ⊢ 𝐻 = (𝑛 ∈ 𝑍 ↦ (𝑚 ∈ ℕ ↦ {𝑥 ∈ dom (𝐹‘𝑛) ∣ -𝑚 < ((𝐹‘𝑛)‘𝑥)})) ⇒ ⊢ (𝜑 → dom 𝐺 ∈ 𝑆) | ||
Theorem | smfinfdmmbl 46209* | If a countable set of sigma-measurable functions have domains in the sigma-algebra, then their infimum function has the domain in the sigma-algebra. This is the fifth statement of Proposition 121H of [Fremlin1] p. 39 . (Contributed by Glauco Siliprandi, 1-Feb-2025.) |
⊢ Ⅎ𝑛𝜑 & ⊢ Ⅎ𝑥𝜑 & ⊢ Ⅎ𝑥𝐹 & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ 𝑍 = (ℤ≥‘𝑀) & ⊢ (𝜑 → 𝑆 ∈ SAlg) & ⊢ (𝜑 → 𝐹:𝑍⟶(SMblFn‘𝑆)) & ⊢ ((𝜑 ∧ 𝑛 ∈ 𝑍) → dom (𝐹‘𝑛) ∈ 𝑆) & ⊢ 𝐷 = {𝑥 ∈ ∩ 𝑛 ∈ 𝑍 dom (𝐹‘𝑛) ∣ ∃𝑦 ∈ ℝ ∀𝑛 ∈ 𝑍 𝑦 ≤ ((𝐹‘𝑛)‘𝑥)} & ⊢ 𝐺 = (𝑥 ∈ 𝐷 ↦ inf(ran (𝑛 ∈ 𝑍 ↦ ((𝐹‘𝑛)‘𝑥)), ℝ, < )) ⇒ ⊢ (𝜑 → dom 𝐺 ∈ 𝑆) | ||
Theorem | sigarval 46210* | 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 46211* | 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 46212* | 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 46213* | 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 46214* | 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 46215* | 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 46216* | 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 46217* | 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 46218* | 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 46219* | 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 46220* | 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 46221* | 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 46222* | 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 46223* | 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 46224* | 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 46225* | 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 46226* | 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 46227 | 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 46228* | 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 46229* |
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 46228 three times and then using cevathlem1 46227 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 46224. 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))) | ||
Theorem | simpcntrab 46230 | 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)) | ||
Theorem | et-ltneverrefl 46231 | Less-than class is never reflexive. (Contributed by Ender Ting, 22-Nov-2024.) Prefer to specify theorem domain and then apply ltnri 11347. (New usage is discouraged.) |
⊢ ¬ 𝐴 < 𝐴 | ||
Theorem | et-equeucl 46232 | Alternative proof that equality is left-Euclidean, using ax7 2012 directly instead of utility theorems; done for practice. (Contributed by Ender Ting, 21-Dec-2024.) |
⊢ (𝑥 = 𝑧 → (𝑦 = 𝑧 → 𝑥 = 𝑦)) | ||
Theorem | et-sqrtnegnre 46233 | 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 46234* | 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 46235* | 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 46236 | Extend class notation to include the set of strictly increasing sequences. |
class UpWord 𝑆 | ||
Definition | df-upword 46237* | 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 46238 | Empty set is an increasing sequence for every range. (Contributed by Ender Ting, 19-Nov-2024.) |
⊢ ∅ ∈ UpWord 𝑆 | ||
Theorem | upwordisword 46239 | Any increasing sequence is a sequence. (Contributed by Ender Ting, 19-Nov-2024.) |
⊢ (𝐴 ∈ UpWord 𝑆 → 𝐴 ∈ Word 𝑆) | ||
Theorem | singoutnword 46240 | 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 46241 | 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 46242 | Singleton is an increasing sequence for any compatible range. (Contributed by Ender Ting, 21-Nov-2024.) |
⊢ 𝐴 ∈ 𝑆 ⇒ ⊢ 〈“𝐴”〉 ∈ UpWord 𝑆 | ||
Theorem | upwordsseti 46243 | 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 46244 | Concatenation of identical singletons is never an increasing sequence. (Contributed by Ender Ting, 22-Nov-2024.) |
⊢ 𝐴 ∈ V ⇒ ⊢ ¬ (〈“𝐴”〉 ++ 〈“𝐴”〉) ∈ UpWord 𝑆 | ||
Theorem | upwrdfi 46245* | 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) | ||
Theorem | hirstL-ax3 46246 | 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 46247 | Recover ax-3 8 from hirstL-ax3 46246. (Contributed by Jarvin Udandy, 3-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.) |
⊢ ((¬ 𝜑 → ¬ 𝜓) → (𝜓 → 𝜑)) | ||
Theorem | aibandbiaiffaiffb 46248 | 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 46249 | 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 46250 | 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 46251 | Given a is equivalent to ⊤, there exists a proof for a. (Contributed by Jarvin Udandy, 30-Aug-2016.) |
⊢ (𝜑 ↔ ⊤) ⇒ ⊢ 𝜑 | ||
Theorem | aisfina 46252 | Given a is equivalent to ⊥, there exists a proof for not a. (Contributed by Jarvin Udandy, 30-Aug-2016.) |
⊢ (𝜑 ↔ ⊥) ⇒ ⊢ ¬ 𝜑 | ||
Theorem | bothtbothsame 46253 | 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 46254 | 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 46255 | 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 46256 | 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 46257 | Given a is exclusive to b, there exists a proof for (not (a if-and-only-if b)); df-xor 1506 is a closed form of this. (Contributed by Jarvin Udandy, 7-Sep-2016.) |
⊢ (𝜑 ⊻ 𝜓) ⇒ ⊢ ¬ (𝜑 ↔ 𝜓) | ||
Theorem | aiffnbandciffatnotciffb 46258 | 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 46259 | 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 46260 | 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 46261 | 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 46262 | 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 46263 | 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 46264 | 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 46265 | 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 46266 | If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.) |
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓)) | ||
Theorem | bisaiaisb 46267 | Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.) |
⊢ ((𝜓 ↔ 𝜑) → (𝜑 ↔ 𝜓)) | ||
Theorem | atbiffatnnbalt 46268 | If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.) |
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓)) | ||
Theorem | abnotbtaxb 46269 | Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.) |
⊢ 𝜑 & ⊢ ¬ 𝜓 ⇒ ⊢ (𝜑 ⊻ 𝜓) | ||
Theorem | abnotataxb 46270 | Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.) |
⊢ ¬ 𝜑 & ⊢ 𝜓 ⇒ ⊢ (𝜑 ⊻ 𝜓) | ||
Theorem | conimpf 46271 | 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 46272 | 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 46273 | 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 46274 | 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 46275 | 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 46276 | 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 46277 | 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 46278 | 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 46279 | 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 46280 | 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 46281 | 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 46282 | 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 46283 | 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 46284 | 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 46285 | 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 46286 | For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.) |
⊢ (𝜑 ∧ 𝜓) ⇒ ⊢ (𝜑 → 𝜓) | ||
Theorem | onenotinotbothi 46287 | 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 46288 | 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 46289 | 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 46290 | 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 46291 | 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 46292 | 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 46293 | 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 46294 | 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 46295 | 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 46296 | 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 46297 | 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 46298 | 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 46299 | 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 46300 | 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 |
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