I'm not exactly sure how to simplify this further.
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1Is that "((not C) or (not (d xor P))) and (C or (d xor P))"? – Beta Oct 11 '13 at 00:19
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It's ((not C) and (not (d xor P))) or (C and (d xor P)) – Ziamor Oct 11 '13 at 00:23
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4This question appears to be off-topic because it is about math, which is probably better handled at math.stackexchange.com. – templatetypedef Oct 11 '13 at 00:24
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1How about "C = d xor P"? – Beta Oct 11 '13 at 00:35
1 Answers
2
So basically your formula is true if C is 1 & d and p are different or if C is 0 & d and p are equal. So when C
is true then d ^ p
must be true, when C
is false d ^ p
must be false. So C
and ~ (d ^ p)
must always be different. (~
is NOT, ^
is XOR, &
is AND)
So it should be equivalent to:
C ^ ( ~ (d ^ p))
which can even be written as
~ (C ^ (d ^ p))
Its truth table should be
C d p d ^ p ~(d ^ p) C ^ (~(d ^ p))
0 0 0 0 1 1
0 0 1 1 0 0
0 1 0 1 0 0
0 1 1 0 1 1
1 0 0 0 1 0
1 0 1 1 0 1
1 1 0 1 0 1
1 1 1 0 1 0
Now compare it to the truth table of your expression:
C d p d ^ p ~ (d ^ p) ~C ~C & ~(d ^ p) C & (d ^ p) ~C & ~(d ^ p) | C & (d ^ p)
0 0 0 0 1 1 1 0 1
0 0 1 1 0 1 0 0 0
0 1 0 1 0 1 0 0 0
0 1 1 0 1 1 1 0 1
1 0 0 0 1 0 0 0 0
1 0 1 1 0 0 0 1 1
1 1 0 1 0 0 0 1 1
1 1 1 0 1 0 0 0 0

Jack
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