Udělejte pravdivou tabulku návrhu itionq [(p (q) V ~ p]?

Udělejte pravdivou tabulku návrhu itionq [(p (q) V ~ p]?
Anonim

Odpovědět:

Viz. níže.

Vysvětlení:

Vzhledem k: #not p -> (p ^^ q) vv ~ p #

Logické operátory:# "ne p:" ne p, ~ p; "a:" ^ ^; nebo: vv #

Logické tabulky, negace:

#ul (| "" p | "" q | "" ~ p | "" ~ q |) #

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

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

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

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

Logické tabulky a & nebo:

#ul (| "" p | "" q | "" p ^^ q "" | "" qvvq "" |) #

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

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

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

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

Logické tabulky, pokud:

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

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

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

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

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

Daná logická část 1:

#ul (| "" p ^^ q "" | "" ~ p "" | "" (p ^^ q) vv ~ p |) #

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

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

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

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

Daná logická část 2:

#ul (| "" ~ q "" | "" (p ^^ q) vv ~ p | "" ~ q -> (p ^^ q) vv ~ p |) #

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

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

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

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