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bsc-leon-vatthauer/thesis/.vscode/ltex.hiddenFalsePositives.en-US.txt

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{"rule":"POSSESSIVE_APOSTROPHE","sentence":"^\\QFurthermore, in mathematical textbooks equality between morphisms is usually taken for granted, i.e. there is some global notion of equality that is clear to everyone.\\E$"}
{"rule":"SENTENCE_WHITESPACE","sentence":"^\\QKleisli').\\E$"}
{"rule":"SENTENCE_WHITESPACE","sentence":"^\\QConstruction.\\E$"}
{"rule":"NO_SPACE_CLOSING_QUOTE","sentence":"^\\Q[inline]Change quotation marks ”\\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q”: Given a Kleisli triple \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, we get a monad \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q where \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q is the object mapping of the Kleisli triple together with the functor action \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q is the morphism family of the Kleisli triple where naturality is easy to show and \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q is a natural transformation defined as \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"IF_IS","sentence":"^\\QLet \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q and \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q be a setoid morphism, we define \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q point wise: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q if \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q if \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q where \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q is defined corecursively by: \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QThe following conditions hold: There exists a unit morphism \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for any DX, satisfying \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q For any \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q there exists a unique morphism \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q satisfying: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q There exists a unique morphism \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q satisfying: &&&& &&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q We will make use of the fact that every \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q is a final coalgebra: [\\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q] This follows by definition of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q.\\E$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QWe call a morphism \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q guarded if there exists an \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q such that the following diagram commutes: &&& &&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QIt suffices to show \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, because then follows: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q We prove this by coinduction using: [inline]Change name of morphism &&& &&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\Q[inline]Make this more explicit The first step in both equations can be proven by monicity of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q and then using \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q and the dual diagram for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q which is a direct consequence of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q: &&&& &&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q the last step holds generally for any \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q.\\E$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QFirst we need to show naturality of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, i.e. we need to show that \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q The needed coalgebra is shown in this diagram: &&&& &&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q Next we check the strength laws: [\\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q] To show that \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q we do coinduction using the following coalgebra: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q [\\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q] We don't need coinduction to show \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, but we need a small helper lemma: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q which is a direct consequence of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q.\\E$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QFirst we need to show naturality of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, i.e. we need to show that \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q The coalgebra for coinduction is: &&&& &&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q We write \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q to abbreviate the rather long coalgebra, i.e. in this diagram \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\Q&&&&&&& &&&&&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q then follows from this diagram.\\E$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QSince \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q is a final coalgebra we get the following proof principle: Given two morphisms \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, to show that \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q it suffices to show that there exists a coalgebra structure \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q such that the following diagrams commute: &&&&&& &&&&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q Uniqueness of the coalgebra morphism \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q then gives us that indeed \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q.\\E$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QFirst we need to show naturality of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, i.e. we need to show that \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q The coalgebra for coinduction is: &&&& &&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q We write \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q to abbreviate the used coalgebra, i.e. in this diagram \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QNext we check the strength laws: [\\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q] To show that \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q we do coinduction using the following coalgebra: &&& &&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q [\\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q] We don't need coinduction to show \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, but we need a small helper lemma: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q which is a direct consequence of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q.\\E$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QIt suffices to show \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, because then follows: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q We prove this by coinduction using: &&& &&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\Q[inline]Move this remark to the beginning of proof The first step in both equations can be proven by monicity of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q and then using \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q and the dual diagram for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q which is a direct consequence of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q: &&&& &&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QTo show that \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q it suffices to show that there exists a coalgebra structure \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q such that the following diagrams commute: &&&&&& &&&&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q Uniqueness of the coalgebra morphism \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q then results in \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q.\\E$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QA (unguarded) Elgot Algebra \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q consists of: An object X for every \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q the iteration \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, satisfying: law:fixpoint Fixpoint: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q Uniformity: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q Folding: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q test: \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QA (unguarded) Elgot Algebra \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q consists of: An object X for every \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q the iteration \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, satisfying: law:fixpoint Fixpoint: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q law:uniformity Uniformity: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q law:folding Folding: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q test: \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QA (unguarded) Elgot Algebra \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q consists of: An object X for every \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q the iteration \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, satisfying: law:fixpoint Fixpoint: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q law:uniformity Uniformity: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q law:folding Folding: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\Qlaw:diamond Diamond\\E$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QA (unguarded) Elgot Algebra \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q consists of: An object X for every \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q the iteration \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, satisfying: law:fixpoint Fixpoint: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q law:uniformity Uniformity: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q law:folding Folding: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\Qlaw:stutter Stutter: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q law:diamond Diamond: \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\Qlaw:stutter Stutter: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q law:diamond Diamond: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q law:compositionality Compositionality: \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QA (unguarded) Elgot Algebra \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q consists of: An object A for every \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q the iteration \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, satisfying: law:fixpoint Fixpoint: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q law:uniformity Uniformity: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q law:folding Folding: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\Qlaw:stutter Stutter: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q law:diamond Diamond: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q law:compositionality Compositionality: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q for \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
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