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`
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`
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`
This material is intended as a summary. Use your textbook for detail explanation.
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