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1. p ∧ ¬q = T

2. (q ∧ p) → r = T

3.¬p → ¬r = T

4.(¬q ∧ p) → r = T

From Eq 1, we got p = T and q = F
Now Apply value of P in Eq 3, we get:
$$\begin{array}{cc} p&¬p&r&¬r&¬p\to r\\ \hline \color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{T}}&\color{red}{\text{F}}&\color{red}{\text{T}}\\ \color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{F}}&\color{red}{\text{T}}&\color{red}{\text{T}}\\ \text{F}&\text{T}&\text{T}&\text{F}&\text{F}\\ \text{F}&\text{T}&\text{F}&\text{T}&\text{T} \end{array}$$

Now there are two possibilities when ¬p→r is T, and ¬p is F but the r has two separate values.

Is this System consistent or inconsistent?

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  • $\begingroup$ If the issue has been resolved, consider accepting and/or upvoting answers: it scores points, signals resolution, and prevents bumping and automatic deletion. $\endgroup$
    – ryang
    Nov 9 at 1:57
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It's perfectly ok that your set of sentences is consistent if you have 2 different models satisfying your set of sentences (a theory) since consistency has nothing to do with uniqueness of model as referenced here

a consistent theory is one that does not lead to a logical contradiction. The lack of contradiction can be defined in either semantic or syntactic terms. The semantic definition states that a theory is consistent if it has a model, i.e., there exists an interpretation under which all formulas in the theory are true.

In fact any set of tautological sentences such as {$(p=p), (q=q), (r=r)$} can always have different truth values for any propositional sentence $p,q,r$ to stay to be consistent.

But look further about your set of particular sentences, $r$ can only be true from the constraint of your last sentence 4 since the antecedent of your material conditional is true then $r$ has to be true...

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A set of sentences as consistent iff their conjunction is satisfiable.

(Informally: a consistent system is one whose premises/axioms are coherent in some universe.)

So, in propositional logic, an inconsistent system is one whose conjunction is a contradiction, i.e., whose conjunction is false regardless of the combination of truth values of its atomic propositions.

So, in your exercise, the system is inconsistent iff $$(1 ∧ 2 ∧ 3 ∧ 4) \equiv\bot,$$ i.e., regardless of $(p,q,r)$'s value, $(1 ∧ 2 ∧ 3 ∧ 4)=$ F,

i.e., every row of $(1 ∧ 2 ∧ 3 ∧ 4)$'s truth table has a False main connective.

Because the main connective $\to$ in your simplified truth table of $(1 ∧ 2 ∧ 3 ∧ 4)$ is True thrice, your system is consistent.

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For a system to be consistent it must have $\textbf{one}$ outcome that is true, $T$. For a system to be inconsistent it would no true values in the outcome, in other words is all false values, $F$.

In your example we see we have $T$'s in the right most column meaning this system $\textbf{is}$ consistent.

$$\begin{array}{c|c} p&¬p&r&¬r&¬p\to r\\ \hline \color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{T}}&\color{red}{\text{F}}&\color{red}{\text{T}}\\ \color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{F}}&\color{red}{\text{T}}&\color{red}{\text{T}}\\ \text{F}&\text{T}&\text{T}&\text{F}&\text{F}\\ \text{F}&\text{T}&\text{F}&\text{T}&\text{T} \end{array}$$

For a system to be in consistent, here is an example: $$\begin{array}{c|c} p & (p\land \lnot p) \\ \hline T & F \\ F &F \end{array}$$ The right most column is always $F$ so this system is inconsistent.

Note: If the right most column is always $T$, that is called a tautology.

Also, for these problems I would do a truth table for each one. You won't have that many rows and it would be easy to see if the system is consistent or not.

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