Faceți masa de adevăr a propoziției ¬q [(pΛq) V ~ p]?

Faceți masa de adevăr a propoziției ¬q [(pΛq) V ~ p]?
Anonim

Răspuns:

Vezi mai jos.

Explicaţie:

Dat: # nu p -> (p ^^ q) vv ~ p #

Operatori logici:# "nu p:" nu p, ~ p; "și:" ^^; sau: vv #

Mese logice, negare:

################################################# #

# "" T | "" T | "" F | "" F | #

# "" T | "" F | "" F | "" T | #

# "" F | "" T | "" T | "" F | #

# "" F | "" F | "" T | "" T | #

Tabele logice și & sau:

################################################################################################## #

# | "" T | "" T | "" T "" | "" T "" |

"|" | | | | | | | | | | | | | | | | | | | | | | | | | | |

# | "" | | | | | | | | | | | | | | | | | | | | | | | | | | |

F | "" F | "" F "" | "" F "" |

Tabele logice, dacă:

#ul (| "" p | "" q | "" p-> q "" |

# | "" T | "" T | "" T "" |

# | "" T | "" F | "" F "" |

# | "" F | "" T | "" T "" |

# | "" F | "" F | "" T "" |

Având în vedere propunerea logică partea 1:

####################################################################### #

"T" "|" "F" "|" "T" "|

# "" F "" | "" F "" | "" F "" |

"T" "|" "" T "" | # "

"T" "|" "" T "" | # "

Având în vedere propunerea logică partea 2:

##########################################################.

"T" "|" "" T "" | # "

| | | | | | | | | | | | | | | | | | | |

"T" "|" "" T "" | # "

"T" "|" "T" "|" "T" "|