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Logic, sets, and recursion

Robert L. Causey

Chapter 1

Sentential Calculus - all with Video Answers

Educators


Chapter Questions

01:08

Problem 1

According to precise and strict application of the syntax rules, which of the following expressions are $\mathcal{S C}$ sentences and which are not?
1. $((P \vee \neg Q) \rightarrow S)$
3. ( $\neg A)$
4. $\neg(\neg A \wedge B)$
5. $(\neg A \leftrightarrow \neg(A \wedge B))$
6. $(\neg(P \neg Q))$
7. $(\neg(\neg P \leftrightarrow(A \vee B)) \rightarrow \neg(A \vee B))$
8. $\left(S_1 \rightarrow R_{13}\right)$
9. $(\phi \rightarrow \psi)$
10. $((Q \rightarrow R) \rightarrow P \vee Q)$

James Kiss
James Kiss
Numerade Educator
06:11

Problem 2

Sketch the tree structure of each of the expressions in the previous exercise that is a sentence. Also, for each of these sentences, list all of its proper subsentences.

Trang Hoang
Trang Hoang
Numerade Educator
00:39

Problem 3

1. The machine runs only if both Switch1 is on and Switch2 is on.
2. For the machine to run, it is necessary that the power cord be plugged in.
3. If a fuse is blown, then the machine does not run.
4. Therefore: If the machine does not run and the power cord is plugged in and Switch1 is on and Switch2 is on, then the fuse is changed.

Mark Scythian
Mark Scythian
Numerade Educator
01:21

Problem 4

1. Part A has failed or part B has failed.
2. If the gadget passes test number one, then it is not the case that part $B$ has failed.
3. If the battery is dead, then the green indicator light is not on.
4. Therefore: The green indicator light is on only if the gadget does not pass test number one.

Aman Gupta
Aman Gupta
Numerade Educator
04:54

Problem 5

1. Bob gets a raise or Bob gets a bonus or Bob gets nothing new.
2. If Bob gets a raise, then it is not the case that either Bob gets a bonus or that he gets nothing new.
3. Bob gets nothing new if and only if profits do not rise.
4. Therefore: If the cost of widget production remains constant, then: If the demand for widgets rises, then profits rise and Bob gets a raise.

Rashmi Sinha
Rashmi Sinha
Numerade Educator
02:54

Problem 6

1. In order for the patient to live, it is necessary that the doctor perform the surgery.
2. The patient lives iff he does not die.
3. The doctor performs the surgery iff there is not a power failure.
4. There is a power failure if the nuclear plant melts down.
5. Therefore: The nuclear plant melts down only if the patient dies.

Colton Wang
Colton Wang
Numerade Educator
04:01

Problem 7

1. The red light was on or the green light was on.
2. If the green light was on, he took path A.
3. If the red light was on, he took path $B$.
4. If he took path $\mathrm{A}$, he encountered a dragon.
5. If he took path $B$, he encountered two doors.
6. If he encountered two doors, then he opened the left door or he opened the right door.
7. If he encountered a dragon, then he lost the game.
8. If he opened the left door, then he lost the game.
9. He did not lose the game.
10. Therefore: The red light was on and he opened the right door.

Suman Saurav Thakur
Suman Saurav Thakur
Numerade Educator
05:33

Problem 8

For each of the following forms of sentences, write a specific $S C$ sentence that is an instance (example) of that form.
1. a negation of an atomic sentence
2. a negation of a conjunction
3. a conditional with an antecedent that is a negation of a disjunction
4. a biconditional with a left side that is also a biconditional and with a right side that is the disjunction of two atomic sentences
5. a conditional that has a consequent that is the negation of another conditional

Rosina Dapaah
Rosina Dapaah
Numerade Educator
01:03

Problem 9

Write the converse and the contrapositive of each of the following.
1. $(\neg(A \vee B) \rightarrow(B \leftrightarrow(C \wedge B)))$
2. $((A \wedge \neg B) \rightarrow \neg \neg(A \vee A))$

Nick Johnson
Nick Johnson
Numerade Educator

Problem 10

Make a truth table for each of the following $S C$ sentences.
1. $\neg((P \rightarrow Q) \rightarrow Q)$
2. $((A \rightarrow(B \rightarrow C)) \leftrightarrow(A \wedge(B \wedge C)))$
3. $(((A \vee \neg B) \wedge((B \vee C) \wedge(\neg A \vee C))) \rightarrow C)$
4. $\neg(((P \rightarrow Q) \rightarrow P) \rightarrow P)$
5. $((A \rightarrow(B \rightarrow C)) \leftrightarrow((A \wedge B) \rightarrow C))$
6. $((D \leftrightarrow E) \leftrightarrow((D \wedge E) \vee(D \vee E)))$

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01:18

Problem 11

The sentence $(P \wedge Q)$ is true (under an interpretation) if both $P$ and $Q$ are true (under this interpretation); it is false otherwise. The sentence $(P \wedge \neg Q)$ is true if $P$ is true and $Q$ is false; it is false otherwise. Corresponding remarks apply to other conjunctions of $P$ or $\neg P$ with $Q$ or $\neg Q$. Also, notice that $(P \wedge Q) \vee(P \wedge \neg Q)$ is true iff either both $P$ and $Q$ are true, or $P$ is true and $Q$ is false. These remarks can be generalized to sentences of three or more sentential letters. Use these observations to write an $S C$ sentence $\phi$ composed of the three letters $P, Q, R$ that has the following properties: (i) It is a disjunction. (ii) Each disjunct of $\phi$ is a conjunction of $P$ or $\neg P$ with a conjunction of $Q$ or $\neg Q$ with $R$ or $\neg R$. (iii) The sentence $\phi$ is true if $P, Q, R$ are all true or $P, Q, R$ are all false; otherwise, it is false. Make a truth table for $\phi$.

Mohamed Mohamed
Mohamed Mohamed
Numerade Educator
01:18

Problem 12

Use the information given in the previous exercise to write an $S C$ sentence $\psi$ that is true if one and only one of $P, Q, R$ is true, and is false otherwise.

Mohamed Mohamed
Mohamed Mohamed
Numerade Educator

Problem 13

One should be a tautology, one an unsatisfiable sentence, and the third a contingent sentence. Make up sentences that are different from the examples in the text. Construct the truth table for each sentence.

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00:52

Problem 14

Give an interpretation that satisfies the following set of sentences, and briefly show the calculations of the truth values of each sentence.
$$
|P \rightarrow \neg Q, Q \wedge R, \neg R \vee Q|
$$

JH
J Hardin
Numerade Educator
00:33

Problem 15

Give an interpretation that satisfies the following set of sentences, and briefly show the calculations of the truth values of each sentence.
$$
\{A \vee(B \rightarrow C), B \wedge \neg D, A \rightarrow D, \neg(E \rightarrow(C \wedge D))\}
$$

Ashley High
Ashley High
Numerade Educator
00:33

Problem 16

Give an interpretation that satisfies the following set of sentences, and briefly show the calculations of the truth values of each sentence.
$$
\{A \wedge(D \vee \neg E), B \rightarrow \neg D, \neg(C \wedge A), C \vee(\neg A \vee B)\}
$$

Ashley High
Ashley High
Numerade Educator

Problem 17

Show that the conclusion of the following argument is not a tautological consequence of the premises:
1. $A \rightarrow B$
2. $C \rightarrow B$
3. $D \rightarrow B$
4. Therefore: $(A \wedge C) \rightarrow D$

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Problem 18

Show that the conclusion of the following argument is a tautological consequence of the premises.
1. $A \rightarrow B$
2. $C \rightarrow B$
3. $B \rightarrow D$
4. Therefore: $(A \wedge C) \rightarrow D$

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01:57

Problem 19

In a murder trial, an expert medical witness testifies, "The physical evidence is consistent with the proposition that the victim was hit on the head with a hammer." Although it is an oversimplification, for the sake of the example, interpret the use of 'consistent' here as tautologically consistent. Explain why the expert's statement does not mean or imply: "The physical evidence (tautologically) implies the proposition that the victim was hit on the head with a hammer."

Victor Salazar
Victor Salazar
Numerade Educator
01:33

Problem 20

Let $\Gamma=\{S, P \leftrightarrow Q, Q \vee R\}$ and $\phi$ be $S \rightarrow \neg R$. 1 . Show that $\phi$ is not a tautological consequence of $\Gamma$. 2. Show that $\Gamma \cup\{\phi \mid$ is satisfiable.

Nick Johnson
Nick Johnson
Numerade Educator

Problem 21

Construct a set of $S C$ sentences with the following properties: It contains three different sentences and it is unsatisfiable. Show that it is unsatisfiable.

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01:33

Problem 22

Let $\Gamma=\{(P \vee Q) \rightarrow R, Q \rightarrow(P \vee R), R \rightarrow Q\}$. Show that $Q$ is tautologically consistent with $\Gamma$ and that $Q$ is tautologically independent of $\Gamma$.

Nick Johnson
Nick Johnson
Numerade Educator
02:23

Problem 23

Use truth tables to show that $\neg(P \wedge Q)$ is tautologically equivalent to $\neg P \vee \neg Q$.

SO
Sigurður Orri
Numerade Educator
03:12

Problem 24

Use truth tables to show that $P \wedge(Q \vee R)$ is tautologically equivalent to $(P \wedge Q) \vee(P \wedge R)$.

Akash M
Akash M
Numerade Educator
04:11

Problem 25

Use truth tables to show that $\neg(A \rightarrow B)$ is tautologically equivalent to $A \wedge \neg B$.

Anthony Ramos
Anthony Ramos
Numerade Educator

Problem 26

Show that
$$
(((A \rightarrow B) \wedge \neg B) \wedge(C \vee A)) \rightarrow(C \vee D)
$$
is tautologous, and thai its converse is not.
$+$

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Problem 27

Let $\Gamma$ and $\Delta$ be sets of $S C$ sentences. Either give a proof or give a counterexample for each of the following statements.
1. If $\Gamma$ is satisfiable and $\Delta$ is satisfiable, then $\Gamma \cup \Delta$ is satisfiable.
2. If $\Gamma \cup \Delta$ is satisfiable, then $\Gamma$ is satisfiable.
A General Instruction: Throughout this book, if an exercise is simply the statement of a theorem or a metatheorem, then the task is to prove this theorem or metatheorem. In your proof, you may use the definitions given in the main text. Unless stated to the contrary, you may also use theorems proved in the text.

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Problem 28

Metatheorem. Let $\tau$ be a tautology and $\Gamma$ be any set of $S C$ sentences. Then $\Gamma \vDash_T \tau$. [See the preceding general instruction.]

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Problem 29

Metatheorem. Let $\rho$ be an $S C$ sentence. Then $\varnothing \vDash_T \rho$ iff $\rho$ is tautologous.

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Problem 30

Let $\rho$ be any $S C$ sentence and let $\tau$ be a tautology. Then $\tau \vDash_T \rho$ iff $\rho$ is tautologous.

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04:19

Problem 31

Let $\phi, \psi, \chi$ be $S C$ sentences. If $\phi \rightarrow \psi, \psi \rightarrow \chi$, and $\chi \rightarrow \phi$ are each tautologous, then each of these sentences is tautologically equivalent to any other one.

Risheek Somu
Risheek Somu
Numerade Educator

Problem 32

Let $\Gamma, \Delta$ be sets of $S C$ sentences, and let $\phi, \psi, \chi$ be $S C$ sentences. Suppose that $\Gamma \vDash_T \phi$ and $\Delta \vDash_T \psi$. Then:
If $\{\phi, \psi\} \vDash_T \chi$, then $\Gamma \cup \Delta \vDash_T \chi$.

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Problem 33

Metatheorem. Let $\Gamma$ be a set of $S C$ sentences and $\phi$ be an $S C$ sentence.
If $\Gamma$ is unsatisfiable, then $(\phi \wedge \neg \phi)$ is a tautological consequence of $\Gamma$, for any $\phi$.

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01:04

Problem 34

Let $\phi, \psi$ be any $\mathcal{S C}$ sentences. Then $(\phi \wedge \neg \phi) \vDash_T \psi$.

Aman Gupta
Aman Gupta
Numerade Educator

Problem 35

Let $\Gamma$ be a set of $S C$ sentences and $\phi$ be an $S C$ sentence, then $\Gamma \vDash_T \phi$ iff $\Gamma \cup\{\neg \phi\}$ is unsatisfiable.

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01:04

Problem 36

Let $\phi_1, \phi_2, \psi$ be $\mathcal{S C}$ sentences. Then:
1. If $\left\{\phi_1, \phi_2 \mid \vDash_T \psi\right.$, then $\phi_1 \rightarrow\left(\phi_2 \rightarrow \psi\right)$ is tautologous; and
2. If $\phi_1 \rightarrow\left(\phi_2 \rightarrow \psi\right)$ is tautologous, then $\left(\phi_1 \wedge \phi_2\right) \rightarrow \psi$ is tautologous; and
3. If $\left(\phi_1 \wedge \phi_2\right) \rightarrow \psi$ is tautologous, then $\left|\phi_1, \phi_2\right| \vDash{ }_T \psi$.

Aman Gupta
Aman Gupta
Numerade Educator
06:53

Problem 37

For each of the following arguments, give an $S C$ derivation of the conclusion from the premise(s). The conclusion is indicated by $1 \therefore$. For instance, in the first problem the conclusion is $C$. Thus, the task is to derive $C$ from the premises, which are $(A \vee B) \rightarrow C$, $R \rightarrow A$, and $R$. The derivation (proof) must be a formal derivation using the format given in Definition 1-17, and using only the $S C$ rules listed in Table 1-2.
1.
$$
\begin{aligned}
& (A \vee B) \rightarrow C \\
& R \rightarrow A \\
& R \quad 1 \therefore C
\end{aligned}
$$
2.
$$
\begin{aligned}
& A \vee \neg(B \wedge C) \\
& / \therefore B \rightarrow(C \rightarrow A)
\end{aligned}
$$
3.
$$
\begin{aligned}
& P \vee Q \\
& \neg(P \wedge A) \\
& \neg Q \quad / \therefore A \rightarrow B
\end{aligned}
$$
4.
$$
\begin{aligned}
& P \rightarrow(Q \vee R) \\
& R \rightarrow \neg P \\
& \neg(P \wedge Q) \quad / \therefore(P \rightarrow S)
\end{aligned}
$$

Bernabe Montoya
Bernabe Montoya
Numerade Educator
00:24

Problem 38

Derive $A \rightarrow(A \vee B)$ from the empty set of premises. (Hint. Assume the antecedent, $A$, as a premise, derive $A \vee B$, then use Rule C.I

Ian Shi
Ian Shi
Numerade Educator
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Problem 39

Derive the following sentence from the empty set of premises: $((P \rightarrow Q) \rightarrow P) \rightarrow P$.

Dmitri Morenike
Dmitri Morenike
Numerade Educator
00:39

Problem 40

The following is an incorrect "derivation" of $A \rightarrow C$ from $B \rightarrow C$. This derivation contains a kind of mistake which is unfortunately rather common. Identify the error in the alleged proof. Also, construct an interpretation which is a counterexample to the claim made by line (6).
$$
\begin{aligned}
& \left\{P r_1\right\} \\
& \left.\mid P_2\right\} \\
& \left\{r_2\right\} \\
& \left\{r_1, P r_2\right\} \\
& \left\{P r_2\right\} \\
& \left.\mid P r_1\right\}
\end{aligned}
$$
(1) $B \rightarrow C$
(2) $A \wedge B$
(3) $B$
(4) $C$
(5) $A$
(6) $A \rightarrow C$
$\mathrm{P}$
$\mathrm{P}$
$\operatorname{Simp}(2)$
MP (3), (1)
$\operatorname{Simp}(2)$
C (5), (4)

Christopher Stanley
Christopher Stanley
Numerade Educator

Problem 41

Derive $K$ from $\Gamma=\{A \rightarrow(B \vee C), \neg B \wedge A, A \rightarrow \neg C\}$. [Hint. $\Gamma$ is unsatisfiable. If you can derive a contradiction from $\Gamma$, e.g., $C$ on a line and $\neg C$ on a line, then any sentence follows by ContraPrm.]

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Problem 42

For each of the following arguments, give an $S C$ derivation of the conclusion from the premise(s). The conclusion is indicated by $/ \therefore$.
1.
$$
\begin{aligned}
& A \vee(B \wedge C) \\
& / \therefore(A \rightarrow C) \rightarrow C
\end{aligned}
$$
2.
$$
\begin{aligned}
& A \rightarrow B \\
& A \vee C \\
& \neg(C \wedge D) \quad / \therefore \quad(A \rightarrow B) \rightarrow(B \vee \neg D)
\end{aligned}
$$
3.
$$
\begin{aligned}
& \neg(A \wedge B) \rightarrow(\neg A \wedge \neg B) \\
& l \therefore A \leftrightarrow B
\end{aligned}
$$
4.
$$
\begin{aligned}
& A \leftrightarrow \neg B \\
& B \vee \neg C \\
& C \rightarrow A \quad / \therefore \quad \neg C
\end{aligned}
$$
5.
$$
\begin{aligned}
& J \rightarrow(K \rightarrow L) \\
& \neg N \rightarrow(J \wedge M) \\
& (K \rightarrow N) \rightarrow \neg P \\
& P \wedge \neg Q \quad / \therefore \quad L \vee Q
\end{aligned}
$$
6.
$$
\begin{aligned}
& \left(A_1 \rightarrow B_1\right) \vee\left(A_2 \rightarrow B_2\right) \\
& l \therefore\left(A_1 \rightarrow B_2\right) \vee\left(A_2 \rightarrow B_1\right)
\end{aligned}
$$
7.
$$
\begin{aligned}
& R \vee(S \vee T) \\
& T \rightarrow P \\
& Q \wedge(\neg P \wedge N) \\
& R \rightarrow S \quad / \therefore S
\end{aligned}
$$
8.
$$
\begin{aligned}
& (F \rightarrow G) \rightarrow H \\
& G \wedge(I \rightarrow J) \\
& \neg(I \wedge \neg J) \rightarrow \neg K \quad / \therefore \quad \neg(H \rightarrow K)
\end{aligned}
$$
9.
$$
\begin{aligned}
& A \vee(B \vee C) \\
& P \rightarrow \neg A \quad / \therefore \quad(P \wedge \neg B) \rightarrow C
\end{aligned}
$$
10.
$$
\begin{aligned}
& A \leftrightarrow B \\
& \neg B \quad 1 \therefore \quad(P \vee A) \rightarrow \neg A
\end{aligned}
$$
11. $\neg C$
$$
(B \rightarrow \neg C) \rightarrow A \quad / \therefore \quad(A \rightarrow C) \rightarrow F
$$
12.
$$
\begin{aligned}
& (P \rightarrow Q) \wedge(Q \rightarrow R) \\
& (R \rightarrow P) \wedge(T \wedge U) \\
& (P \leftrightarrow R) \rightarrow S \\
& (T \wedge S) \rightarrow(V \vee W) \quad / \therefore \quad(U \wedge V) \vee(U \wedge W)
\end{aligned}
$$

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01:45

Problem 43

For each of the following arguments, either give an $S C$ derivation of the conclusion from the premises, or give a counterexample interpretation. If you construct an interpretation, show the truth values of the sentences under this interpretation.
1. $A \vee B$
$$
C \rightarrow \neg A \quad / \therefore C \rightarrow B
$$
2. $(P \vee Q)$
$$
\begin{aligned}
& (P \rightarrow R) \\
& (Q \rightarrow S) \quad / \therefore \quad T \rightarrow(R \vee S)
\end{aligned}
$$
3.
$$
\begin{aligned}
& A \vee E \\
& A \rightarrow((P \vee Q) \rightarrow C) \\
& \neg E \vee C \\
& (A \wedge \neg E) \rightarrow F \quad l \therefore \quad C \vee P
\end{aligned}
$$
4.
$$
\begin{aligned}
& E \rightarrow(F \vee G) \\
& G \rightarrow\left(H_1 \wedge I\right) \\
& \neg H_1 \quad / \therefore \quad(E \rightarrow I)
\end{aligned}
$$
5.
$$
\begin{aligned}
& P \rightarrow(Q \vee R) \\
& (R \rightarrow P) \\
& \neg(P \wedge Q) \quad / \therefore \quad(P \rightarrow S)
\end{aligned}
$$

Amy Jiang
Amy Jiang
Numerade Educator
01:24

Problem 44

Derive each of the following from the empty set of premises:
$$
\begin{aligned}
& ((A \vee \neg A) \wedge B) \leftrightarrow B, \\
& ((A \wedge \neg A) \vee B) \leftrightarrow B .
\end{aligned}
$$

Vysakh M
Vysakh M
Numerade Educator
05:45

Problem 45

A list of selected $S C$ theorems is given at the end of Section 1.6.3. Derive each of these theorems from the empty set of premises. [Remark. Since there are several theorems, this exercise requires several derivations. Some of these derivations are rather long.]

Subhadeepta Sahoo
Subhadeepta Sahoo
Numerade Educator
01:24

Problem 46

Suppose that we wish to achieve some goal $A$. Suppose that, if we achieve $B$ and $C$ and $D$, then we will also achieve $A$. One can also think of $A$ as some problem to solve, where $A$ has the property that it will be solved if each of $B, C$, and $D$ is solved. When this relation holds, it is said that the goal or problem $A$ can be reduced to the subgoals or subproblems $B, C$, and $D$. In sentential calculus, this situation can be represented by:
$$
(B \wedge(C \wedge D)) \rightarrow A .
$$
We can also represent this situation by the tree in Figure 1-2. In this figure, $B, C, D$ are the child nodes of $A$. The curved arc indicates that $A$ is reducible to the conjunction of $B, C, D$.
Figure 1-2.
Now suppose that we achieve $B$ if we achieve $E$, so:
$$
E \rightarrow B .
$$
Suppose that we will also achieve $B$ if we achieve $F$, so:
$$
F \rightarrow B .
$$
We need $B$ and $C$ and $D$ to get $A$, but either child node $E$ or $F$ will get us $B$. This situation is represented in Figure 1-3, where there is no arc between the lines from $B$ to $E$ and from $B$ to $F$.Also suppose that $G$ is a "good" node, i.e., something that we can simply do (or a problem that we can directly solve), without depending on any further action (or further solutions). In sentential calculus, we represent this by:
$G$.
Suppose that $H$ is a "nogood" node, i.e., something that we cannot achieve or realize or solve; it is a dead end. In sentential calculus, we represent this by:
$\neg H$.
In the tree diagram, terminal nodes that are good are represented by * and terminal nodes that are nogood are represented by \#. The situation is now represented by Figure 1-4.
Figure 1-4.
Suppose we have the following additional information:
$$
\begin{gathered}
(I \wedge S) \rightarrow F \\
I \\
(J \wedge K) \rightarrow S \\
J \\
K \\
L \rightarrow C \\
M \rightarrow C \\
T \rightarrow C \\
\neg L \\
M
\end{gathered}
$$
\begin{gathered}
T \\
(N \wedge(O \wedge P)) \rightarrow D \\
N \\
O \\
Q \rightarrow P \\
R \rightarrow P \\
U \rightarrow P \\
\neg Q \\
R \\
U
\end{gathered}
FIGURE CANT COPY

Manik Pulyani
Manik Pulyani
Numerade Educator

Problem 47

Transform the following sentences into Conjunctive Normal Form. Use the informal procedure described in Section 1.7. Cancel out tautologies (as described in the section), and simplify expressions of the forms $\phi \vee \phi$ and $\phi \wedge \phi$ by using the Tautology Laws.
1. $(A \rightarrow B) \rightarrow(C \wedge A)$
2. $\neg(P \rightarrow \neg(Q \vee(\neg R \wedge S)))$

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Problem 48

Let $\tau$ be a tautology and $\phi$ be any $S C$ sentence. Show that it is always possible to construct an $S C$ derivation of
$$
(\tau \wedge \phi) \leftrightarrow \phi
$$
from the empty set of premises. [Hint. Suppose that $(\tau \wedge \phi)$ is placed on the first line of a derivation by Rule P. Then one can infer $\phi$ on the second line by Simp. Continuing in this manner, describe a general schema of a derivation of $(\tau \wedge \phi) \leftrightarrow \phi$ from $\varnothing$. You may assume Completeness; specifically, Metatheorem 1-27.]

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Problem 49

Let $\Gamma=\{A \rightarrow(C \vee D), B \rightarrow A, D \rightarrow C, B, \neg(B \wedge C)\}$. Show that $\Gamma$ is unsatisfiable by deriving a contradiction from it, i.e., construct a derivation with some sentence $\phi$ on one line and with $\neg \phi$ on another line, or else with their conjunction on a line.

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Problem 50

Let the premises of an argument be
$$
\Gamma=\{A \rightarrow(C \vee D), B \rightarrow A, D \rightarrow C, B\},
$$
and let the conclusion be $(B \wedge C)$. Put each premise and the negation of the conclusion into CNF, and construct a Resolution Proof for this argument.

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00:54

Problem 51

Use Resolution to derive a contradiction from the following:
$$
\begin{aligned}
& (A \vee \neg B) \vee C \\
& B \vee \neg D \\
& \neg C \vee D \\
& B \vee(C \vee D) \\
& \neg A \vee \neg B \\
& \neg D \vee \neg B
\end{aligned}
$$

Michelle Nguyen
Michelle Nguyen
Numerade Educator