PHIL 220. Introduction to Logic
We now put translation to work on complex natural language arguments.
Three steps, applied to each argument in turn:
My client is not the burglar. For several reasons. If there are no signs of forced entry, then my client is the burglar only if she had an accomplice inside. But we can discard that option, as no one in the building could have known my client before today. Nor are there any signs of forced entry.
We specify a translation key:
\(p\): There are signs of forced entry.
\(q\): My client is the burglar.
\(r\): My client had an accomplice inside.
\(p\): There are signs of forced entry.
\(q\): My client is the burglar.
\(r\): My client had an accomplice inside.
\(p\): There are signs of forced entry.
\(q\): My client is the burglar.
\(r\): My client had an accomplice inside.
\(p\): There are signs of forced entry.
\(q\): My client is the burglar.
\(r\): My client had an accomplice inside.
\(p\): There are signs of forced entry.
\(q\): My client is the burglar.
\(r\): My client had an accomplice inside.
\(p\): There are signs of forced entry.
\(q\): My client is the burglar.
\(r\): My client had an accomplice inside.
If the mind and the brain are one and the same thing, then the mind is material only if the brain is. But the mind is not material unless it is divisible. The brain is material, but the mind is not divisible. So the mind and the brain are not one and the same thing.
We specify a translation key:
\(p\): The mind and the brain are one and the same thing.
\(q\): The mind is material.
\(r\): The brain is material.
\(t\): The mind is divisible.
\(p\): The mind and the brain are one and the same thing.
\(q\): The mind is material.
\(r\): The brain is material.
\(t\): The mind is divisible.
\(p\): The mind and the brain are one and the same thing.
\(q\): The mind is material.
\(r\): The brain is material.
\(t\): The mind is divisible.
\(p\): The mind and the brain are one and the same thing.
\(q\): The mind is material.
\(r\): The brain is material.
\(t\): The mind is divisible.
\(p\): The mind and the brain are one and the same thing.
\(q\): The mind is material.
\(r\): The brain is material.
\(t\): The mind is divisible.
\(p\): The mind and the brain are one and the same thing.
\(q\): The mind is material.
\(r\): The brain is material.
\(t\): The mind is divisible.
There are \(2^4 = 16\) assignments.
A full truth table would require 16 rows …
We now employ the search for a counterexample method:
Focus on assignments that make the conclusion false
Ask which of the assignments that remain make the premises true.
If we find a counterexample, then the argument is invalid.
Otherwise, the argument is valid.
To make the conclusion \(\neg p\) false, we must have \(p: T\):
\[ p : T \]
So, we need only look at 8 rows out the 16:
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline p & q & r & t &p \to (q \to r) & \neg t \to \neg q & r \wedge \neg t & \neg p \\ \hline T & T & T & T & & & & \color{red}{F} \\ T & T & T & F & & & & \color{red}{F} \\ T & T & F & T & & & & \color{red}{F} \\ T & T & F & F & & & & \color{red}{F} \\ T & F & T & T & & & & \color{red}{F} \\ T & F & T & F & & & & \color{red}{F} \\ T & F & F & T & & & & \color{red}{F} \\ T & F & F & F& & & & \color{red}{F} \\ \hline \end{array} \]
To make the third premise \(r \wedge \neg t\) true, we consider only rows at which
\[ r : T \qquad t : F \] This narrows down the field of potential counterexamples to two rows:
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline p & q & r & t &p \to (q \to r) & \neg t \to \neg q & r \wedge \neg t & \neg p \\ \hline T & T & T & F & & & \color{red}{T} & \color{red}{F} \\ T & F & T & F & & & \color{red}{T} & \color{red}{F} \\ \hline \end{array} \]
To make the second premise \(\neg t \to \neg q\) true, we rule out the row at which \(t\) is false and \(q\) is true:
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline p & q & r & t &p \to (q \to r) & \neg t \to \neg q & r \wedge \neg t & \neg p \\ \hline T & F & F & T & & \color{red}{T} & \color{red}{T} & \color{red}{F} \\ \hline \end{array} \]
The row that remains makes the first premise true, which means that it is a counterexample to validity:
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline p & q & r & t &p \to (q \to r) & \neg t \to \neg q & r \wedge \neg t & \neg p \\ \hline T & T & F & T & \color{red}{T} & \color{red}{T} & \color{red}{T} & \color{red}{F} \\ \hline \end{array} \] It follows that the argument is invalid.
Your insurer will cover your expenses only if you are at fault or are current with your monthly payments. But if you are current with your monthly payments, then you can afford to cover the expenses yourself. So, your insurer will not cover your expenses unless you can afford to cover them yourself.
We specify a translation key:
\(p\): Your insurer will cover your expenses.
\(q\): You are at fault.
\(r\): You are current with your monthly payments.
\(t\): You can afford to cover the expenses yourself.
\(p\): Your insurer will cover your expenses.
\(q\): You are at fault.
\(r\): You are current with your monthly payments.
\(t\): You can afford to cover the expenses yourself.
\(p\): Your insurer will cover your expenses.
\(q\): You are at fault.
\(r\): You are current with your monthly payments.
\(t\): You can afford to cover the expenses yourself.
\(p\): Your insurer will cover your expenses.
\(q\): You are at fault.
\(r\): You are current with your monthly payments.
\(t\): You can afford to cover the expenses yourself.
\(p\): Your insurer will cover your expenses.
\(q\): You are at fault.
\(r\): You are current with your monthly payments.
\(t\): You can afford to cover the expenses yourself.
There are \(2^4 = 16\) assignments.
A full truth table would require 16 rows …
We now employ the search for a counterexample method:
Focus on assignments that make the conclusion false
Ask which of the assignments that remain make the premises true.
If we find a counterexample, then the argument is invalid.
Otherwise, the argument is valid.
To make the conclusion \(\neg t \to \neg p\) false, we must have:
\[ t : F \qquad p : T \]
So, we need look only at 4 rows out the 16:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline p & q & r & t &p \to (q \vee r) & r \to t & \neg t \to \neg p \\ \hline T & T & T & F & & & \color{red}{F} \\ T & T & F & F & & & \color{red}{F} \\ T & F & T & F & & & \color{red}{F} \\ T & F & F & F& & & \color{red}{F} \\ \hline \end{array} \]
To make the third premise \(r \to t\) true, we rule out rows at which
\[ r : T \qquad t : F \] This narrows down the field of potential counterexamples to two rows:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline p & q & r & t &p \to (q \vee r) & r \to t & \neg t \to \neg p \\ \hline T & T & F & F & & \color{red}{T} & \color{red}{F} \\ T & F & F & F & & \color{red}{T} & \color{red}{F} \\ \hline \end{array} \]
To make the first premise \(p \to (q \vee r)\) true, we need only make \(q\) true, which means that the first row is now a counterexample to validity:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline p & q & r & t &p \to (q \vee r) & r \to t & \neg t \to \neg p \\ \hline T & T & F & F & \color{red}{T} & \color{red}{T} & \color{red}{F} \\ \hline \end{array} \] So, we found a counterexample to validity in the assignment: \[ p : T \qquad q : T \qquad r : F \qquad r : F \]
It follows that the argument is invalid.
Here is an argument adapted from W. O. Quine’s Methods of Logic.
If Jones is ill or Smith is away, then unless Robinson comes to his senses, neither will the Argus deal be concluded nor will the directors declare a dividend. Consequently, if Smith is away and Robinson does not come to his senses, the Argus deal will not be concluded.
We specify a translation key:
\(p\): Jones is ill.
\(q\): Smith is away.
\(r\): Robinson comes to his senses.
\(t\): The Argus deal will be concluded.
\(s\): The directors will declare a dvidend.
\(p\): Jones is ill.
\(q\): Smith is away.
\(r\): Robinson comes to his senses.
\(t\): The Argus deal will be concluded.
\(s\): The directors will declare a dvidend.
\(p\): Jones is ill.
\(q\): Smith is away.
\(r\): Robinson comes to his senses.
\(t\): The Argus deal will be concluded.
\(s\): The directors will declare a dvidend.
\(p\): Jones is ill.
\(q\): Smith is away.
\(r\): Robinson comes to his senses.
\(t\): The Argus deal will be concluded.
\(s\): The directors will declare a dvidend.
There are \(2^5 = 32\) assignments.
A full truth table would require 32 rows!
We now employ the search for a counterexample method:
Focus on assignments that make the conclusion false
Ask which of the assignments that remain make the premises true.
If we find a counterexample, then the argument is invalid.
Otherwise, the argument is valid.
To make the conclusion \((q \wedge \neg r) \to \neg t\) false, we must have:
\[ q : T \qquad r : F \qquad t : T \]
So, we need look only at 4 rows out the 32:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline p & q & r & t & s & (p \vee q) \to (\neg r \to (\neg t \wedge \neg s) ) & (q \wedge \neg r) \to \neg t \\ \hline T & T & F & T & T & & \color{red}{F} \\ F & T & F & T & T & & \color{red}{F} \\ T & T & F & T & F & & \color{red}{F} \\ F & T & F & T & F & & \color{red}{F} \\ \hline \end{array} \]
We want to make the first premise \((p \vee q) \to (\neg r \to (\neg t \wedge \neg s) )\) true. Unfortunately:
So, the first premise
\[ (p \vee q) \to (\neg r \to (\neg t \wedge \neg s) ) \]
comes out false in all rows that remain.
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline p & q & r & t & s & (p \vee q) \to (\neg r \to (\neg t \wedge \neg s) ) & (q \wedge \neg r) \to \neg t \\ \hline T & T & F & T & T & F & \color{red}{F} \\ F & T & F & T & T & F & \color{red}{F} \\ T & T & F & T & F & F & \color{red}{F} \\ F & T & F & T & F & F & \color{red}{F} \\ \hline \end{array} \]
It follows that the argument is valid.