OnProgrammingforCorrectness/3_.tex at master · kammitama5/OnProgrammingforCorrectness · GitHub

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\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}

LaTex document for LAFF class notes
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Weakening/ Strengthening
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Homework 1.5.2.1
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For each of the following, if applicable, indicate which statement is TRUE (by examination):
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1. $0 \leq x \leq 10$ is weaker than 1 $\leq x < 5.$
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2. $x = 5 \land y = 4$ is stronger than $y = 4$.
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3. $ x \leq$ 5 $\vee$ $ y = 3 $ is weaker than $x = 5 \land y = 4$.
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4. TRUE and FALSE.
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TRUE is weaker than FALSE.
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Homework 1.5.2.2-1
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Use the Basic Equivalences to prove that:
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E1 $\land$ E2 $\Rightarrow$ E1.
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E1 $\land$ E1 $\Rightarrow$ E1
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$<implication>$
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$\neg$(E1 $\land$ E2) $\vee$ E1
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$<DeMorgan's>$
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$(\neg E1 \vee \neg E2) \vee E1$
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$<commutativity: associativity: commutativity>$
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$\neg E2 \vee (E1 \vee \neg E1)$
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$<excluded middle>$
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$\neg E2 \vee T$
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Homework 1.5.2.2-2
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$E1 \Rightarrow E1 \vee E3$
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$<implication>$
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$\neg E1 \vee (E1 \vee E3)$
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$<associativity: commutativity>$
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$(E1 \vee \neg E1) \vee E3$
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$<excluded middle>$
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$T \vee E3$
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$<commutativity: \vee-simplification>$
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$T$
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Homework 1.5.2.2-3
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Use the Basic Equivalence to prove that:
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$E1 \land E2 \Rightarrow E1 \vee E3$.
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$<implication>$
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$\neg(E1 \land E2) \vee (E1 \vee E3)$
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$<DeMorgan's>$
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$(\neg E1 \vee \neg E2) \vee (E1 \vee E3)$
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$<commutativity: associativity:commutativity x 2>$
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$\neg E2 \vee E3 \vee (E1 \vee \neg E1)$
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$<excluded middle>$
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$\neg E2 \vee E3 \vee T$
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$< \vee - simplification x 2>$
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$T$
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Homework 1.5.2.3
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Use the Basic Equivalences and/or the results from Homework 1.5.2.2 to prove that:
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$E1 \land E2 \Rightarrow (E1 \vee E3) \land E2.$
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$< \land distributivity >$
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$E1 \land E2 \Rightarrow (E1 \land E2) \vee (E3 \land E2)$
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$< p \Rightarrow q \vee q>$
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$T$
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Homework 1.5.2.4
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Consider the three Weakening/Strengthening Laws:
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Weakening/Strengthening:
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$((E1 \land E2) \Rightarrow E1) \Leftrightarrow T$
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$(E1 \Rightarrow (E1 \vee E2)) \Leftrightarrow T$
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$((E1 \land E2) \Rightarrow (E2 \vee E3)) \Leftrightarrow T$
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This exercise shows that if you only decide to remember one, it should be the last one.
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To show that $(E1 \land E2) \Rightarrow E1$ is a special case of $(E1 \land E2) \Rightarrow (E1 \vee E3)$ set $E3$ to :
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FALSE
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To show that $E1 \Rightarrow (E1 \vee E3)$ is a special case of $(E1 \land E2) \Rightarrow (E1 \vee E3)$ set $E2$ to :
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TRUE
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\end{document}