Question

The Logic Problem, taken from WFF’N PROOF, The Game of Logic, has these two assumptions:1. “Logic is difficult or not many students like logic.”2. “If mathematics is easy, then logic is not difficult.”By translating these assumptions into statements involving propositional variables and logical connectives, determine whether each of the following are valid conclusions of these assumptions:a) That mathematics is not easy, if many students like logic.b) That not many students like logic, if mathematics is not easy.c) That mathematics is not easy or logic is difficult.d) That logic is not difficult or mathematics is not easy.e) That if not many students like logic, then either mathematics is not easy or logic is not difficult.

Short answer

(a) Mathematics is not easy, if many students like logic and (d) Logic is not difficult or mathematics is not easy are valid conclusions of the given assumptions.

Step-by-step solution

Important Laws

Logical Equivalences: \(p \to q \equiv (\neg p \vee q)\)and \((p \vee q) \equiv \neg p \to q\)

Hypothetical Syllogism: \(((p \to q) \wedge (q \to r)) \to (p \to r)\)

Conjunction Law: \(((p) \wedge (q)) \to (p \wedge q)\)

De Morgan’s law: \(\neg (p \wedge q) \equiv (\neg p \vee \neg q)\)

Assign propositional variables

Let p be the proposition - “Logic is difficult”

Let q be the proposition - “Many students like logic”

Let r be the proposition – “Mathematics is easy”

The given 2 assumptions (premises) can now be written as:

(1)\((p \vee \neg q)\)

(2) \(r \to \neg p\)

Part (a)

“Mathematics is not easy, if many students like logic” can be written as\(q \to \neg r\), to check this, the following steps are there:

Step Reason

1. \((p \vee \neg q)\) Premise
2. \((\neg q \vee p)\) Commutative law on (1)
3. \(q \to p\) Logical equivalence of (2)
4. \(r \to \neg p\) Premise
5. \(p \to \neg r\) Contrapositive of (4)
6. \(((q \to p) \wedge (p \to \neg r)) \to (q \to \neg r)\) Hypotherical Syllogism from (3) & (5)

Hence\(q \to \neg r\)is true, (a) is a valid conclusion

Part (b)

“Not many students like logic, if mathematics is not easy” can be written as \(\neg r \to \neg q\).

Take contrapositive of \(q \to \neg r\)(proved in Step 3), \(r \to \neg q\)

This is not logically equivalent to \(\neg r \to \neg q\). Hence, conclusion (b) is invalid.

Part (c)

“Mathematics is not easy or logic is difficult” can be represented as \((\neg r \vee p)\).

Premise \((r \to \neg p)\)

Its logical equivalence can be given as \((\neg r \vee \neg p)\), which is not logically equivalent with \((\neg r \vee p)\).

Hence, conclusion (c) is also invalid.

Part (d)

“Logic is not difficult or mathematics is not easy” is written as \((\neg p \vee \neg r)\), which is the logical equivalence of the premise \((r \to \neg p)\).

Hence conclusion (d) is valid.

Part (e)

“If not many students like logic, then either mathematics is not easy or logic is not difficult” can be written as\(\neg q \to (\neg p \vee \neg r)\) or \(q \to (p \wedge r)\).

Step Reason

1. \((\neg q \vee \neg r)\) Logical Equivalence of (6)
2. \(((\neg q \vee p) \wedge (\neg q \vee \neg r))\) Conjunction from (2) & (7)
3. \(\neg q \vee (p \wedge \neg r)\) Distribtive law on (8)
4. \(q \to (p \wedge \neg r)\) Logical Equivalence from (9)
5. \(\neg q \to (\neg p \vee r)\) De Morgan’s Law on (10)

Here\(\neg q \to (\neg p \vee r)\)is not logically equivalent to\(\neg q \to (\neg p \vee \neg r)\).

Hence, conclusion (e) is invalid.

Thus, onlyconclusions (a) and (d)are valid.