Erstellt von Freddy Ulate Agüero
vor fast 10 Jahre
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Frage | Antworten |
Simplificación (Simp.) | \[ \underline{1. \,\, P \vee Q \,\,} \\ \therefore P, Q \] |
Adjunción (Adj.) | \[1. \,\, P \,\,\,\,\,\,\,\,\,\,\, \\ \underline{2. \,\, Q \,\,\,\,\,\,\,\,\,\,\,} \\ \therefore P \wedge Q \] |
Adición (Adi.) | \[ \underline{1. \,\, P \,\,\,\,\,\,\,\,\,\,\,\,} \\ \therefore P \vee Q \] |
Modus Pones (MP) | \[1. \,\, P \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \underline{2. \,\, P \rightarrow Q} \\ \therefore Q \] |
Modus Tollens (MT) | \[1. \,\, \neg Q \,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \underline{2. \,\, P \rightarrow Q \,\,\,} \\ \therefore \neg P \] |
Silogismo disyuntivo (SD) | \[1. \,\, \neg P \,\,\,\,\,\,\,\, \\ \underline{2. \,\, P \wedge Q } \\ \therefore Q \] |
Silogismo Hipotético (SH) | \[1. \,\, P \rightarrow Q \,\,\,\,\, \\ \underline{2. \,\, Q \rightarrow R \,\,\,\,\,} \\ \therefore P \rightarrow R \,\, \] |
Dilema Constructivo (DC) | \[1. \,\,\,\, P \vee Q \,\,\,\,\, \\ 2. \,\, P \rightarrow Q \,\,\,\,\, \\ \underline{3. \,\, Q \rightarrow S \,\,\,\,\,} \\ \therefore P \vee R \,\,\,\, \] |
Dilema Destructivo (DD) | \[1. \,\, \neg R \vee \neg S \,\,\, \\ 2. \,\,\,\, P \rightarrow R \,\,\,\,\,\, \\ \underline{3. \,\,\,\, Q \rightarrow S \,\,\,\,\,\,\,} \\ \therefore \neg Q \vee \neg S \] |
Ley de Casos (LC) | \[1. \,\,\,\,\,\, P \rightarrow Q \\ \underline{2. \,\, \neg P \rightarrow R } \\ \therefore Q \vee R \] |
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