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7 part 2
|}
Thus we can see that #1, 2, and 4 are equivalent to p.q⊃r while #3 is equivalent to p∨q⊃r
{|
! p || q || r || r⊃p.q || r⊃p∨q || (r⊃p).(r⊃q) || (r⊃p)∨(r⊃q)
|-
|T || T || T || T || T || T || T
|-
|T || T || F || T || T || T || T
|-
|T || F || T || F || T || F || T
|-
|T || F || F || T || T || T || T
|-
|F || T || T || F || T || F || T
|-
|F || T || F || T || T || T || T
|-
|F || F || T || F || F || F || F
|-
|F || F || F || T || T || T || T
|}
Thus we can see that #1 is equivalent to r⊃p.q and #2 is equivalent to r⊃p∨q
8.