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


(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 $Val_{\mathfrak{M},s}(\theta)=1$


$(Val_{\mathfrak{M},s}(\varphi)=1$ 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 ($\neq 0$)?


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 :


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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.