Question: some are valid and some are invalid ( 1 is valid ) ( 1 ) AAxP ( x ) : . AAx [ P (

some are valid and some are invalid (1 is valid)
(1) AAxP(x)
:.AAx[P(x)Q(x)]
:.AAxQ(x)
(2) AAxP(x)
:.AAx[P(x)Q(x)]
:.EExQ(x)
(4) AAx F,
:.EEx[P(x)Q(x)]
:.EExQ(x)
(6) EExP(x)
:.AAx[P(x)Q(x)]
:.EEEE(x)
(7) EExP(x)
EEx[P(x)Q(x)]
:.AAxQ(x)
(8),EExP(x)
,EEx[P(x)Q(x)]
:.EExQ(x)
(9) AAxnotQ(x)
AAx[P(x)Q(x)]
:.AAxnotP(x)
(10) AAxnotQ(x)
:.AAx[P(x)Q(x)]
:.EExnotP(x)
(11){:[AAxnotQ(x)]:.EEx[P(x)Q(x)]:.AAxnotP(x)EExnotQ(x)(15)EEx[P(x)Q(x)]:.AAxnotP(x)
]):}
[:.xnotP(x)
(16) EExnotQ(x)
EEx[P(x)Q(x)]
:.EExnotP(x)
pick some valid syllogism(s) above, and prove their validity.
Recall that that for a syllogism with two premises P1 and P2 and one conclusion C, its validity is shown by proving (P1 And P2)=>C.
leverage the four inference rules introduced we need to operate in predicate logic.
leverage the four inference rules introduced and mind the restrictions. Kemember to specify the number of each valid svllogism.
Valid syllogism 1(Number: )
Valid syllogism 2(Number: )
 some are valid and some are invalid (1 is valid) (1)

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