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Propositional calculus - Prove or refute - Not a contradiction Could anyone help me with this? Prove or refute: ![enter image description here]( What is the definition of $\nvdash_{CPL}$? And what does it mean about valuation $V$? How can I prove that there is (or isn't) valuation $V$ such that $V(A)=\mathrm{true}$?

$\
vdash_{\text{CPL}}$ means : " **not** derivable in Classical Propositional Logic.

We can use the Completeness Theorem : if $\vDash$, then $\vdash$.

Thus, assume that $A$ is a contradiction. This means that, for every _valuation_ $v$, we have $v(A)= \text F$.

Thus, there is no valuation $v$ such that $v(A)$ is TRUE **and** $v(B→C)$ is FALSE (irrespective of the truth-value assigned by $v$ to formula $B \to C$).

This, means that $A \vDash B → C$, because a contradiction entails every formula.

Thus, by Completeness : $A \vdash B \to C$.

Contraposinf the argument, we have that :

> if **not** $A \vdash B \to C$, then $A$ is **not** a _contradiction_.

* * *

Another approach is through the Proof system : the details depend on the specific proof system used for $\text{CPL}$.

Basically, we have the EFQ rule :

> $\dfrac {\bot}{\varphi}$.

Thus, using $B \to C$ as $\varphi$, if $A$ is a contradiction, we would have :

> $A \vdash_{\text{CPL}} B \to C$,

contrary to assumption.

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