Implication

1. p => q = (not p) or q

Solution:
To prove `p=>q=~pvvq`, we have to first prepare the following truth table

`(1)`	`(2)`	`(3)=(1)=>(2)`	`(4)=~(1)`	`(5)=(4)vv(2)`
`p`	`q`	`p=>q`	`~p`	`~pvvq`
T	T	T	F	T
T	F	F	F	F
F	T	T	T	T
F	F	T	T	T

from this table, we can say that columns (3) and (5) are identical.
`:. p=>q=~pvvq`

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2. p => q = (not q) => (not p)

Solution:
To prove `p=>q=~q=>~p`, we have to first prepare the following truth table

`(1)`	`(2)`	`(3)=(1)=>(2)`	`(4)=~(2)`	`(5)=~(1)`	`(6)=(4)=>(5)`
`p`	`q`	`p=>q`	`~q`	`~p`	`~q=>~p`
T	T	T	F	F	T
T	F	F	T	F	F
F	T	T	F	T	T
F	F	T	T	T	T

from this table, we can say that columns (3) and (6) are identical.
`:. p=>q=~q=>~p`

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3. not(p => q) = p and (not q)

Solution:
To prove `~(p=>q)=p^^~q`, we have to first prepare the following truth table

`(1)`	`(2)`	`(3)=(1)=>(2)`	`(4)=~(3)`	`(5)=~(2)`	`(6)=(1)^^(5)`
`p`	`q`	`p=>q`	`~(p=>q)`	`~q`	`p^^~q`
T	T	T	F	F	F
T	F	F	T	T	T
F	T	T	F	F	F
F	F	T	F	T	F

from this table, we can say that columns (4) and (6) are identical.
`:. ~(p=>q)=p^^~q`

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