Not exactly sure how I can further simplify these logic gates [closed]

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I'm not exactly sure how to simplify this further.

Answer 1

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