next up previous
Next: Prove or Disprove the Up: bool-ans Previous: bool-ans

Simplify as much as possible.

  1. $(AB + C) (AB + C')$
    $=AB(C+C')$ (distributive law (16))
    $=AB(1)$ (identity 7)
    $=AB$ (identity 2)
    A \& B

  2. $X'W + XW' + XW + X'W'$
    $=X'W'+X'W+XW'+XW$ (commutative law (10))
    $=X'(W'+W)+X(W'+W)$ (distributive law)
    $=(X'+X)(W'+W)$ (distributive law)
    $=1\cdot1$ (identity 7)
    $=1$ (identity 3)
  3. $pqr + rp$
    $=p(qr+r)$ (distributive)
    $=p(q+1)r$ (distributive)
    $=pr$ (identity 3)
    p \& r

  4. $(y'+z')xyz$
    $=xyy'z+xyzz'$ (distributive)
    $=0+0=0$ (identity 8)
  5. $(C+DX) (C+EX)$
    $=C+(DX)(EX)$ (distributive law (15))
    $=C+DEX$ (identity 6)
    C | (D \& E \& X)
  6. $T(L'+V) + TV'$
    $=T(L'+V+V')$ (distributive law)
    $=T(L'+1)=T$ (identities 7 and 3)
  7. $fg \oplus 1$
    $=0fg+1(fg)'$ (definition)
    $=f'+g'$ (DeMorgan's law)
    ~f | ~g

  8. $(m \oplus n) \oplus (m' \oplus n)$
    $=(m \oplus m') \oplus (n \oplus n)$ (XOR is commutative & associative)
    $=1 \oplus 0=1$ (identities)
  9. $(uw \oplus (t+u))'$
    $=((uw)'(t+u)+uw(t+u)')'$ (definition)
    $=((u'+w')(t+u)+uu'wt)'$ (DeMorgan's)
    $=(u'+w')'+(t+u)'$ (identities & DeMorgan's)
    $=uw+t'u'$ (DeMorgan's once again)
    u \& w | ~t \& ~u
  10. $(xy+z)'(z(x'+y'))$
    $=z'(xy)'z(x'+y')=0$ (DeMorgan's and identity)
  11. $[(x+a)(x+b)(x+c)(x+d)]'$
    $=(x+abcd)'=x'(abcd)'$ (Distributive & DeMorgan's)
    ~x \& ~(a \& b \& c \& d)



MM Hugue 2008-01-27