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i think this is invalid.
P=T, Q=T, R=T.

All the premises are true but the conclusion is false. So, the argument is invalid.
Not valid.  It is a contingency

so , The question is (p->q)  ∧  (q->r) → r’

so now , we try to make  arguement false if happens then not valid else valid.

the premises gives p->r (transistivity) and now we make our conclusion r’ false , which means r is true .

hence it makes premise true . so , NOT VALID.

A formula is valid if it is true for any values of its variables.

Conversely, a formula is not valid, if there exists at least one combination of values of variables for which it is false.

We can use either method to solve for validity.

Case 1: Trying to prove tautology.

$[(p \rightarrow q) \land (q \rightarrow r)] \rightarrow r’$

$=[(p’ \lor q) \land (q’ \lor r)] \rightarrow r’$

$=[(p’ \lor q) \land (q’ \lor r)]’ \lor r’$

$=(p \land q’) \lor (q \land r’) \lor r’$

Re-writing the connectives using +, *, ‘ for simplicity sake:

$=pq’ + qr’ + r’$

$=pq’ + r’$

which is NOT a tautology.

Hence, the formula is NOT valid.

Case 2: Trying to find atleast one combination where the formula yields false.

An implication $A \rightarrow B$ is false, iff $A=True$ and $B=False$

$[(p \rightarrow q) \land (q \rightarrow r)] \rightarrow r’$

Let’s assume, $RHS = False$

i.e., $r’ = False \Rightarrow r = True$

$LHS:$

$[(p \rightarrow q) \land (q \rightarrow r)]$

$=(p’ + q) (q’ + r)$

$=p’q’ + p’r + qr$

Putting the value of $r=True \text{, or } 1$

$=p’q’ + p’.1 + q.1$

$=p’q’ + p’ + q$

$=p’ + q$

$=p \rightarrow q$

which is false, if $p=True$ and $q=False$

Since, when $RHS$ is false, $LHS$ can also become false, the original formula is NOT valid.

### 1 comment

This can be solved if one remembers the Laws of Inference – Hypothetical Syllogism.

$[(p \rightarrow q) \land (q \rightarrow r)] \rightarrow (p \rightarrow r)$

which is a tautology, hence valid.

$\therefore [(p \rightarrow q) \land (q \rightarrow r)] \rightarrow r’$ cannot be valid.