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K-map and universal gate: xy + x'z + yz

Short answer

F = xy + x'z (yz is redundant). Realized with NAND gates: double inversion gives F = ((xy)'.(x'z)')'. Circuit uses four NANDs: one for xy, one inverter (NAND) for x', one for x'z, and one to combine.

Simplify the following Boolean expression using K-map and realize it by using universal gate of your interest. F(x, y, z) = xy + x'z + yz

F(x,y,z) = xy + x'z + yz.

Minterms: xy = m6 + m7, x'z = m1 + m3, yz = m3 + m7. So F = Σm(1, 3, 6, 7).

K-map

          yz
        00  01  11  10
 x = 0 | 0 | 1 | 1 | 0 |
 x = 1 | 0 | 0 | 1 | 1 |

Groups: (m6, m7) = xy, and (m1, m3) = x'z. The group (m3, m7) = yz is redundant (it is the consensus term).

Simplified: F = xy + x'z

Realization using NAND gates

Using double inversion: F = ((xy + x'z)')' = ((xy)' · (x'z)')'

 x ------------+
               [NAND 1]--- (xy)' ------+
 y ------------+                       |
                                       [NAND 4]--- F = xy + x'z
 x ---[NAND 2]--- x' --+               |
     (inputs tied)     [NAND 3]--- (x'z)'
 z --------------------+
  • NAND 1: NAND(x, y) = (xy)'
  • NAND 2 (used as NOT, inputs joined): gives x'
  • NAND 3: NAND(x', z) = (x'z)'
  • NAND 4: NAND of the NAND 1 and NAND 3 outputs gives F = xy + x'z

Total: 4 two-input NAND gates (the universal gate used).

✓ Answer checked by Claude Opus 5.5 (high effort)

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