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The best way to get a proper understanding is to write down the truth table completely. You have 3 inputs (A, B, S), so this will give you 2\$^3\$ = 8 combinations:

S  A  B | S' C  D | Z
--------+---------+--
0  0  0 | 1  0  0 | 0
0  0  1 | 1  0  0 | 0
0  1  0 | 1  1  0 | 1
0  1  1 | 1  1  0 | 1
1  0  0 | 0  0  0 | 0
1  0  1 | 0  0  1 | 1
1  1  0 | 0  0  0 | 0
1  1  1 | 0  0  1 | 1

It's often useful to add intermediate results to make things more clear. I added a term C = (A \$\land\$ S') and D = (B \$\land\$ S). Now it should be clear that Z = (C \$\lor\$ D).

The best way to get a proper understanding is to write down the truth table completely. You have 3 inputs (A, B, S), so this will give you 2\$^3\$ = 8 combinations:

S  A  B | S' C  D | Z
--------+---------+--
0  0  0 | 1  0  0 | 0
0  0  1 | 1  0  0 | 0
0  1  0 | 1  1  0 | 1
0  1  1 | 1  1  0 | 1
1  0  0 | 0  0  0 | 0
1  0  1 | 0  0  1 | 1
1  1  0 | 0  0  0 | 0
1  1  1 | 0  0  1 | 1

It's often useful to add intermediate results to make things more clear. I added a term \$C = (A \land S')\$ and \$D = (B \land S)\$. Now it should be clear that \$Z = (C \lor D)\$.

The best way to get a proper understanding is to write down the truth table completely. You have 3 inputs (A, B, S), so this will give you 2\$^3\$ = 8 combinations:

S  A  B | S' C  D | Z
--------+---------+--
0  0  0 | 1  0  0 | 0
0  0  1 | 1  0  0 | 0
0  1  0 | 1  1  0 | 1
0  1  1 | 1  1  0 | 1
1  0  0 | 0  0  0 | 0
1  0  1 | 0  0  1 | 1
1  1  0 | 0  0  0 | 0
1  1  1 | 0  0  1 | 1

It's often useful to add intermediate results to make things more clear. I added a term C = (A \$\land\$ S') and D = (B \$\land\$ S). Now it should be clear that Z = (C \$\lor\$ D).

The best way to get a proper understanding is to write down the truth table completely. You have 3 inputs (A, B, S), so this will give you 2\$^3\$ = 8 combinations:

S  A  B | S' C  D | Z
--------+---------+--
0  0  0 | 1  0  0 | 0
0  0  1 | 1  0  0 | 0
0  1  0 | 1  1  0 | 1
0  1  1 | 1  1  0 | 1
1  0  0 | 0  0  0 | 0
1  0  1 | 0  0  1 | 1
1  1  0 | 0  0  0 | 0
1  1  1 | 0  0  1 | 1

It's often useful to add intermediate results to make things more clear. I added a term C = (A \$\land\$ S') and D = (B \$\land\$ S). Now it should be clear that Z = (C \$\lor\$ D).

The best way to get a proper understanding is to write down the truth table completely. You have 3 inputs (A, B, S), so this will give you 2\$^3\$ = 8 combinations:

S  A  B | S' C  D | Z
--------+---------+--
0  0  0 | 1  0  0 | 0
0  0  1 | 1  0  0 | 0
0  1  0 | 1  1  0 | 1
0  1  1 | 1  1  0 | 1
1  0  0 | 0  0  0 | 0
1  0  1 | 0  0  1 | 1
1  1  0 | 0  0  0 | 0
1  1  1 | 0  0  1 | 1

It's often useful to add intermediate results to make things more clear. I added a term C = (A \$\land\$ S') and D = (B \$\land\$ S). Now it should be clear Z = (C \$\lor\$ D).

The best way to get a proper understanding is to write down the truth table completely. You have 3 inputs (A, B, S), so this will give you 2\$^3\$ = 8 combinations:

S  A  B | S' C  D | Z
--------+---------+--
0  0  0 | 1  0  0 | 0
0  0  1 | 1  0  0 | 0
0  1  0 | 1  1  0 | 1
0  1  1 | 1  1  0 | 1
1  0  0 | 0  0  0 | 0
1  0  1 | 0  0  1 | 1
1  1  0 | 0  0  0 | 0
1  1  1 | 0  0  1 | 1

It's often useful to add intermediate results to make things more clear. I added a term C = (A \$\land\$ S') and D = (B \$\land\$ S). Now it should be clear that Z = (C \$\lor\$ D).