Show $\models(\phi\rightarrow(\psi\rightarrow\theta))\rightarrow((\phi\rightarrow\psi)\rightarrow(\phi\rightarrow\theta))$

Question

Question: Show $\models(\phi\rightarrow(\psi\rightarrow\theta))\rightarrow((\phi\rightarrow\psi)\rightarrow(\phi\rightarrow\theta))$

Answer:

(1) Let, $\mathfrak{M},s\models(\varphi\rightarrow(\psi\rightarrow\theta)\rightarrow((\varphi\rightarrow\psi)\rightarrow(\varphi\rightarrow\theta))$

$\Leftrightarrow Val_{\mathfrak{M},s}(\varphi\rightarrow(\psi\rightarrow\theta))=0$ or $Val_{\mathfrak{M},s}((\varphi\rightarrow\psi)\rightarrow(\varphi\rightarrow\theta))=1$

(2) Suppose $Val_{\mathfrak{M},s}(\varphi\rightarrow(\psi\rightarrow\theta))\neq 0$ (i.e. $=1$)

Then, $Val_{\mathfrak{M},s}(\varphi\rightarrow(\psi\rightarrow\theta))=1$

$\Leftrightarrow Val_{\mathfrak{M},s}(\varphi)=0$ or $Val_{\mathfrak{M},s}(\psi\rightarrow\theta))=1$

$\Leftrightarrow Val_{\mathfrak{M},s}(\varphi)=0$ or $Val_{\mathfrak{M},s}(\psi)=0$ or $Val_{\mathfrak{M},s}(\theta)=1$

(3) Then, $Val_{\mathfrak{M},s}((\varphi\rightarrow\psi)\rightarrow(\varphi\rightarrow\theta))=1$

$\Leftrightarrow Val_{\mathfrak{M},s}((\varphi\rightarrow\psi))=0$ or $Val_{\mathfrak{M},s}(\varphi\rightarrow\theta)=1$

$\Leftrightarrow (Val_{\mathfrak{M},s}(\varphi)=1$ and $Val_{\mathfrak{M},s}(\psi)=0)$ or $Val_{\mathfrak{M},s}(\varphi)=0$ or $Val_{\mathfrak{M},s}(\theta)=1$

(4) So,

$Val_{\mathfrak{M},s}(\varphi)=0$ or $Val_{\mathfrak{M},s}(\psi)=0$ or $Val_{\mathfrak{M},s}(\theta)=1$

implies

$(Val_{\mathfrak{M},s}(\varphi)=1$ and $Val_{\mathfrak{M},s}(\psi)=0)$ or $Val_{\mathfrak{M},s}(\varphi)=0$ or $Val_{\mathfrak{M},s}(\theta)=1$

Point of contention:

I understand all of the steps, but dont understand why the conclusion (4) is necessarily true. Also in step (2), why do we assume the statement is true ($\neq 0$)?

Answer 1

For (2) we have apply "at the meta-level" the equivalence between $p \lor q$ and $\lnot p \to q$.

Thus, "$Val(φ→(ψ→θ))=0$ or $Val((φ→ψ)→(φ→θ))=1$" is equivalent to : "if $Val(φ→(ψ→θ)) \ne 0$, then $Val((φ→ψ)→(φ→θ))=1$".

Then the proof goes one "unwinding" separately the antecedent :

$Val(φ→(ψ→θ)) \ne 0$

in (2) to derive the equivalent condition :

(2a) $ \ Val(φ)=0$ or $Val(ψ)=0$ or $Val(θ)=1$

and the consequent :

$Val((φ→ψ)→(φ→θ))=1$

in (3) to derive the equivalent condition :

(3a) $ \ (Val(φ)=1$ and $Val(ψ)=0)$ or $Val(φ)=0$ or $Val(θ)=1$.

Now, going back to the initial conditional, we can use (2a) and (3a) above to express it as :

if $[ \ Val(φ)=0$ or $Val(θ)=1$ or $Val(ψ)=0 \ ]$, then $[ \ Val(φ)=0$ or $Val(θ)=1$ or $(Val(φ)=1$ and $Val(ψ)=0) \ ]$.

To show that this implication holds, we can argue by cases :

(i) if $Val(φ)=0$ holds, then "$Val(φ)=0$ or ..." holds;

(ii) the same if $Val(θ)=1$.

The "tricky case" is :

(iii) if $Val(ψ)=0$ holds; in this case we have two sub-cases : $Val(φ)=0$ holds, and thus again : "$Val(φ)=0$ or ..." holds; or $Val(φ)=1$ holds, and thus "$(Val(φ)=1$ and $Val(ψ)=0)$" holds, and so also "$Val(φ)=0$ or ..." holds.

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Comment

It seems to me a very complicated way to "describe with words" an eight-lines truth-table, i.e. a quite simple one ...