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Chapter 2: Problem 17

Beweisen Sie die Abtrennregel (modus ponens): $$ (A \wedge(A \Rightarrow B)) \Rightarrow B $$

Short Answer

Expert verified
Short Answer: To prove the Modus Ponens rule, we redefined the implications and applied a series of logical laws, including the distribution law, contradiction simplification, identity law, exportation law, and tautology. After these transformations, we showed that a true proposition A and a true conditional statement (A implies B) result in B being true, validating Modus Ponens.

Step by step solution

01

Define the rule

The rule we want to prove, Modus Ponens, can be expressed as: \( (A \wedge (A \Rightarrow B)) \Rightarrow B \)
02

Apply the definition of implication

Recall that any implication \(A \Rightarrow B\) can be represented as \((\lnot A \vee B)\). Therefore, we can rewrite our rule as: \( (A \wedge (\lnot A \vee B)) \Rightarrow B \)
03

Distributive Law

Apply the distributive law on the left side of the implication. Distribute '\(\wedge\)' over '\(\vee\)' to get: \( ((A \wedge \lnot A) \vee (A \wedge B)) \Rightarrow B \)
04

Contradiction simplification

Notice that \((A \wedge \lnot A)\) is a contradiction, meaning it is always FALSE. Thus, we can simplify the expression to: \( (False \vee (A \wedge B)) \Rightarrow B \)
05

Identity law

We know that anything OR'ed with False, results in the original statement. This is called the identity law, which states that \(A \vee False = A\). So, our implication becomes: \( (A \wedge B) \Rightarrow B \)
06

Exportation law

According to the exportation law, \((A \wedge B) \Rightarrow C\) is equivalent to \(A \Rightarrow (B \Rightarrow C)\). Using this law here, we obtain: \( A \Rightarrow (B \Rightarrow B) \)
07

Tautology

\(B \Rightarrow B\) is always true, a tautology, so we can write: \( A \Rightarrow True \) Now, it's clear that if we provide true propositions A and (A implies B), B will also be true. Therefore, we have successfully proved the Modus Ponens rule, \( (A \wedge (A \Rightarrow B)) \Rightarrow B \).

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Most popular questions from this chapter

Beweisen Sie die 脛quivalenzen: $$ \begin{aligned} (A \vee B) & \Leftrightarrow \neg(\neg A \wedge \neg B) \\ (A \wedge B) & \Leftrightarrow \neg(\neg A \vee \neg B) \\ (A \Rightarrow B) & \Leftrightarrow((\neg A) \vee B) \end{aligned} $$

Welche der folgenden Aussagen sind richtig? F眉r alle \(x \in \mathbb{R}\) gilt: 1.,\(x>1\) ist hinreichend f眉r \(x^{2}>1 . "\) 2.,\(x>1\) ist notwendig f眉r \(x^{2}>1 .\) 3.,\(x \geq 1\) ist hinreichend f眉r \(x^{2}>1\)." 4.,\(x \geq 1\) ist notwendig f眉r \(x^{2}>1 .\)

Jene reellen Zahlen \(x\), die L枚sung einer Polynomgleichung $$ a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}=0 $$ mit Koeffizienten \(a_{k} \in \mathbb{Z}\) sind, nennt man algebraische Zahlen. Dabei muss mindestens ein \(a_{k} \neq 0\) sein. Alle rationalen Zahlen sind algebraisch, aber auch viele irrationale Zahlen geh枚ren zu dieser Klasse, etwa \(\sqrt{2}\). Reelle Zahlen, die nicht algebraisch sind, hei脽en transzendent. Zeigen Sie, dass unter der Voraussetzung, dass jedes Polynom nur endlich viele Nullstellen hat (was wir bald ohne M眉he beweisen werden k枚nnen), die Menge aller algebraischen Zahlen abz盲hlbar ist.

Es seien \(M_{1}\) und \(M_{2}\) Teilmengen von \(X\). Beweisen Sie die einfachste Form der Regeln von de Morgan, wobei wir \(C_{X}\) als Bezeichnung f眉r die Komplementbildung bez眉glich \(X\) verwenden: $$ \begin{aligned} &C_{X}\left(M_{1} \cap M_{2}\right)=C_{X}\left(M_{1}\right) \cup C_{X}\left(M_{2}\right) \\ &C_{X}\left(M_{1} \cup M_{2}\right)=C_{X}\left(M_{1}\right) \cap C_{X}\left(M_{2}\right) \end{aligned} $$ Stellen Sie diesen Sachverhalt mittels Venn-Diagrammen dar.

Die Menge \(A_{4}\) hat vier Elemente, die Mengen \(B_{3}\), \(B_{4}\) und \(B_{5}\) haben entsprechend drei, vier und f眉nf Elemente. 脺berlegen Sie jeweils, ob es Abbildungen $$ \begin{aligned} &f_{43}: A_{4} \rightarrow B_{3} \\ &f_{44}: A_{4} \rightarrow B_{4} \\ &f_{45}: A_{4} \rightarrow B_{5} \end{aligned} $$ geben kann, die (a) injektiv, aber nicht surjektiv, (b) surjektiv, aber nicht injektiv, (c) bijektiv sind.
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