Options

Exercise 7 - Chapter 4

edited June 2018 in Exercises

Show that \( \Rightarrow \) as defined in Eq. (4.6) indeed satisfies Eq. (4.5).

Previous Next

$$ \tag{4.5} b \wedge c \le d \text{ iff } b \le (c \Rightarrow d) $$ $$ \tag{4.6} \begin{array}{c c | c} c & d & c \Rightarrow d \\ \hline \text{true} & \text{true} & \text{true} \\ \text{true} & \text{false} & \text{false} \\ \text{false} & \text{true} & \text{true} \\ \text{false} & \text{false} & \text{true} \end{array} $$

Comments

  • 1.
    edited June 2018

    Here I show that it is true via a truth table.

    $$ \begin{array}{c c c | c c | c c | c } b & c & d & b \wedge c & ( b \wedge c ) \le d & c \Rightarrow d & b \le ( c \Rightarrow d ) & ( b \wedge c ) \le d = b \le ( c \Rightarrow d ) \\ \hline \text{true} & \text{true} & \text{true} & \text{true} & \text{true} & \text{true} & \text{true} & \text{true} \\ \text{true} & \text{true} & \text{false} & \text{true} & \text{false} & \text{false} & \text{false} & \text{true} \\ \text{true} & \text{false} & \text{true} & \text{false} & \text{true} & \text{true} & \text{true} & \text{true} \\ \text{true} & \text{false} & \text{false} & \text{false} & \text{true} & \text{true} & \text{true} & \text{true} \\ \text{false} & \text{true} & \text{true} & \text{false} & \text{true} & \text{true} & \text{true} & \text{true} \\ \text{false} & \text{true} & \text{false} & \text{false} & \text{true} & \text{false} & \text{true} & \text{true} \\ \text{false} & \text{false} & \text{true} & \text{false} & \text{true} & \text{true} & \text{true} & \text{true} \\ \text{false} & \text{false} & \text{false} & \text{false} & \text{true} & \text{true} & \text{true} & \text{true} \\ \end{array} $$ A more interesting approach would support the notion that they are equal because Bool is a quantale.

    Comment Source:Here I show that it is true via a truth table. \[ \begin{array}{c c c | c c | c c | c } b & c & d & b \wedge c & ( b \wedge c ) \le d & c \Rightarrow d & b \le ( c \Rightarrow d ) & ( b \wedge c ) \le d = b \le ( c \Rightarrow d ) \\\\ \hline \text{true} & \text{true} & \text{true} & \text{true} & \text{true} & \text{true} & \text{true} & \text{true} \\\\ \text{true} & \text{true} & \text{false} & \text{true} & \text{false} & \text{false} & \text{false} & \text{true} \\\\ \text{true} & \text{false} & \text{true} & \text{false} & \text{true} & \text{true} & \text{true} & \text{true} \\\\ \text{true} & \text{false} & \text{false} & \text{false} & \text{true} & \text{true} & \text{true} & \text{true} \\\\ \text{false} & \text{true} & \text{true} & \text{false} & \text{true} & \text{true} & \text{true} & \text{true} \\\\ \text{false} & \text{true} & \text{false} & \text{false} & \text{true} & \text{false} & \text{true} & \text{true} \\\\ \text{false} & \text{false} & \text{true} & \text{false} & \text{true} & \text{true} & \text{true} & \text{true} \\\\ \text{false} & \text{false} & \text{false} & \text{false} & \text{true} & \text{true} & \text{true} & \text{true} \\\\ \end{array} \] A more interesting approach would support the notion that they are equal because **Bool** is a quantale.
Sign In or Register to comment.