Can one deduce a formula from a set of $\mathcal{L}$-formulas with wrong values!?

Question

Please have look at this propositional consequence problem (page 54, A Friendly Introduction to Mathematical Logic):

$\Gamma = \{P(x,y,x), x<y ~\vee M(\omega,p), \neg P(x,y,z) ~\wedge (\neg x < y)\}$

I'm gonna investigate into the deduction of $\neg M(\omega,p)$ from set of $\mathcal{L}$-formulas above.

1- AFAIK, all formulas of $\Gamma$ must be true. So please have a look at below deduction:

$P(x,y,x)$ is true. So $\neg P(x,y,x)$ is false and its logical intersection with every other formula is false, too. So, the third element of $\Gamma$, i.e. $\neg P(x,y,z) ~\wedge (\neg x < y)$ is false. How is it possible?

2- The other ambiguity is about the precedence of operators where both logical and algebraic operators will be used, together. As an instance, $(x < y)$ is something usual to compare $x$ and $y$, but what does $(\neg x < y)$ mean?

Answer 1

You have introduced a typo; it is :

$\Gamma = \{ P(x,y,x), x < y \lor M(w,p), \lnot P(x,y,x) \land (\lnot x < y) \}$.

We have to "map" the formulae to propositional formulae, replacing atoms with propositional variables:

$\Gamma^* = \{ A, B \lor C, \lnot A \land \lnot B \}$

and $\lnot C$ for $\lnot M(w,p)$.

Thus, we have reduced the problem to check if $\Gamma^* \vDash \lnot C$.

So, your first comment is correct: the set $\Gamma^*$ (and obviously also $\Gamma$) is inconsistent : no truth assignment can makes all the formulas in it simultaneously true.

This means that there is no truth assignment that makes each formulas in $\Gamma^*$ true and makes $\lnot M(w,p)$ false.

Conclusion, $\lnot M(w,p)$ is (vacuously) a propositional consequence (see Def.2.4.1) of $\Gamma$.

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See Principle of explosion (or Ex falso (sequitur) quodlibet) :

from contradiction, anything (follows).