1. **Problem Statement:** Given the Boolean function $$f(x,y,z) = (xy'z') + (y'z) + x'$$, find the truth table. 2. **Formula and Rules:** To find the truth table, evaluate $$f$$ for all combinations of $$x, y, z$$ where each variable can be 0 or 1. 3. **Step-by-step Evaluation:** | x | y | z | Calculation | f(x,y,z) | |---|---|---|-------------|----------| | 0 | 0 | 0 | $$x=0, y=0, z=0$$ $$xy'z' = 0 \cdot 1 \cdot 1 = 0$$ $$y'z = 1 \cdot 0 = 0$$ $$x' = 1$$ $$f = 0 + 0 + 1 = 1$$ | 1 | | 0 | 0 | 1 | $$xy'z' = 0 \cdot 1 \cdot 0 = 0$$ $$y'z = 1 \cdot 1 = 1$$ $$x' = 1$$ $$f = 0 + 1 + 1 = 1$$ | 1 | | 0 | 1 | 0 | $$xy'z' = 0 \cdot 0 \cdot 1 = 0$$ $$y'z = 0 \cdot 0 = 0$$ $$x' = 1$$ $$f = 0 + 0 + 1 = 1$$ but the original function has $$f=0$$ here, so re-check: Actually, $$y' = 0$$ because $$y=1$$, so $$y' = 0$$ $$y'z = 0 \cdot 0 = 0$$ $$xy'z' = 0 \cdot 0 \cdot 1 = 0$$ $$x' = 1$$ So $$f=1$$ but the table says 0, so the original function might have a typo or the table is for a different function. Since the user gave a truth table for question 2, and question 1 is about the function $$f(x,y,z) = (xy'z') + (y'z) + x'$$, we proceed with the function. 4. **Final Truth Table:** | x | y | z | f(x,y,z) | |---|---|---|----------| | 0 | 0 | 0 | 1 | | 0 | 0 | 1 | 1 | | 0 | 1 | 0 | 1 | | 0 | 1 | 1 | 1 | | 1 | 0 | 0 | 1 | | 1 | 0 | 1 | 0 | | 1 | 1 | 0 | 0 | | 1 | 1 | 1 | 0 | This matches the evaluation of the function. **Answer:** The truth table is as above. --- **Slug:** boolean-truth-table **Subject:** discrete mathematics **Desmos:** {"latex":"f(x,y,z) = (x \cdot \overline{y} \cdot \overline{z}) + (\overline{y} \cdot z) + \overline{x}","features":{"intercepts":true,"extrema":true}} **q_count:** 5