I READ A new introduction to modal logic at page 176-177
it said in modal logic, irreflexive frame preserves the one rule, i.e.
Gabb.
and proof's content is as follow :
I am having trouble understanding from the point after the statement that the consequent of the Gabb rule has value 0 at w1.
Why does it follow that if there is a world where the consequent of the Gabb rule is not valid, then it implies that the wff index and world index coincide,
and why does the last sentence is deduced from the second to last sentence?
I shall trace Hughes and Cresswell's argument with some explicative notes. The argument is remarkably instructive in a broader perspective.
First, preliminary notes: A modal logical system $\mathbf{S}$ is said to be characterised by a class $\mathbf{C}$ of frames if $\mathbf{S}$ is sound and complete with respect to every frame $\mathcal{F}\in\mathbf{C}$.
A formula $\theta$ characterises a class $\mathbf{C}$ of frames $\iff\mathcal{F}\Vdash\theta$, for all $\mathcal{F}\in\mathbf{C}$.
For example, the formula $\Box p\rightarrow p$ (i.e., axiom $\mathrm{T}$) characterises the class of reflexive frames. In other words, the system $\mathbf{T}$ is characterised by the class of reflexive frames.
Updating the notation a bit, we have what they call Gabb rule (with axiom $\mathrm{T}$ as a component):
$$\vdash\alpha_{1}\rightarrow\Box(\alpha_{2}\rightarrow\Box(\alpha_{3}\rightarrow\ldots\Box(\alpha_{n}\rightarrow(\underbrace{\Box p\rightarrow p}_{axiom\, \mathrm{T}}))\ldots)$$ $$\implies\vdash\alpha_{1}\rightarrow\Box(\alpha_{2}\rightarrow\Box(\alpha_{3}\rightarrow\ldots\Box\neg\alpha_{n}))\ldots)$$
where $p$ does not occur in any of $\alpha_{1},\ldots,\alpha_{n}$. Formally, it would be appropriate to call this statement a metatheorem (on irreflexiveness) rather than a rule in the given context, however, the authors seem to introduce a notion of “characterising rule” to link it up to the notion of characterising formula, for there is no formula that corresponds to all and only irreflexive frames.
For ease of reference, let us rewrite the metatheorem as
$$\vdash\alpha_{1}\rightarrow\phi\implies\vdash\alpha_{1}\rightarrow\psi$$
Assume that there is an irreflexive frame $\mathcal{F}_{irr}=\langle W, R_{irr}\rangle$, that is, no frame is related to itself as regards accessibility (note that this is a stronger condition than non-reflexiveness):
$$\forall w(w\in W\rightarrow \neg wR_{irr}w)$$
on which the formula $\alpha_{1}\rightarrow\psi$ does not hold. Therefore, there is a model $\mathcal{M}_{irr}$ for which $\alpha_{1}\rightarrow\psi$ is not satisfiable:
$$\mathcal{M}_{irr}\not\Vdash\alpha_{1}\rightarrow\psi$$
Hence, for some $w_{1}\in W$ and valuation function $V$
$$V(\alpha_{1}\rightarrow\psi, w_{1})=0$$
A material implication is false only when its antecedent is true and its consequent is false. Therefore, at $w_{1}$, $\alpha_{1}$ must be true and $\psi$ must be false. Consider $\psi$ in the form of $\Box\psi'$:
$$\Box(\underbrace{\alpha_{2}\rightarrow\Box(\alpha_{3}\rightarrow\ldots\Box\neg\alpha_{n}))\ldots}_{\psi'})$$
In order for $\psi$ to be false at $w_{1}$, by the truth condition of $\Box$, $\psi'$ must be false at least one world $w_{2}\in W$ such that $w_{1}R_{irr}w_{2}$.
Hence, as in the preceding step, $\alpha_{2}$ must be true and the consequent composed of nested implications must be false. Iterating the evaluation in this fashion, we reach the last component
$$\Box(\alpha_{n-1}\rightarrow\Box\neg\alpha_{n})$$
where $\alpha_{n-1}$ must be true at $w_{n-1}$ and $\Box\neg\alpha_{n}$ must be false. Then, $\alpha_{n}$ must be true at some $w_{n}$ so that $\neg\alpha_{n}$ is false.
Thereby, we obtain a chain of worlds $w_{1},\ldots, w_{n}$ and a sequence of propositions $\alpha_{1},\ldots,\alpha_{n}$ such that $V(\alpha_{k}, w_{k})=1,\; k=1,\ldots, n$.
Now, consider another model $\mathcal{M}_{irr}^\ast$ such that
\begin{equation} V^\ast(p, w_{k})=\begin{cases} 1, & \text{if $k\neq n$}.\\ 0, & \text{if $k=n$}. \end{cases} \end{equation}
Since $p$ does not occur in the sequence of propositions $\alpha_{1},\ldots,\alpha_{n}$ (thus, does not interfere with their truth-values), the propositions in the sequence $k=1,\ldots,n$
$$V^\ast(\alpha_{k},w_{k}) = V(\alpha_{k},w_{k}) = 1$$
At $w_{n}$, $p$ is false, while at all other $w_{k}$ accessible from $w_{n}$ is true. Notice that since the frame $\mathcal{F}_{irr}$ is irreflexive, $w_{n}$ is not accessible from $w_{n}$, thus, falsity at $w_{n}$ does not interfere with the truth-value of $\Box p$ and it is true. Consequently, the formula $\Box p\rightarrow p$ is false:
$$V^{\ast}(\Box p\rightarrow p, w_{n}) = 0$$
Let us look at $w_{1}$ the case of $\alpha_{1}\rightarrow\phi$. To swiftly see what to assign to $V^{\ast}(\alpha_{1}\rightarrow\phi)$, we consider the difference between $\psi$ and $\phi$: There is $\alpha_{n}\rightarrow(\Box p\rightarrow p)$ in place of $\neg\alpha_{n}$ evaluated at $w_{n}$. We have already found that $\alpha_{n}$ is true and $\Box p\rightarrow p$ is false. Then, by the “domino effect” of the last component as explained above, we get
$$V^\ast(\alpha_{1}\rightarrow\phi, w_{1})=0$$
Notice that this is an argument by contraposition in the metalanguage. $\alpha_{1}\rightarrow\phi$ is a theorem of system $\mathbf{T}$. When $\alpha_{1}\rightarrow\psi$ is invalidated on an irreflexive frame, the $\mathbf{T}$ theorem is invalidated as a consequence. It is thus concluded that, in a certain sense described by the authors, the “rule” corresponds to irreflexive frames.