0

Definition
We say that a set of propositions (the "premises") $Γ$ entails a proposition $φ$ (the "conclusion") if for every truth assignment $A$, if $A(φ)$=1 when we have $A(ψ)$=1 for all $\psi$ in $Γ$.

Note: if a truth assignment assigns "false" (i.e. 0) to a premise in $Γ$, it doesn't matter what it assigns to $φ$; this truth assignment still satisfies the definition.

The note below makes me confuse. Does it mean if there exists $A(ψ)$=0 for $\psi$ in $Γ$, $Γ$ entails $φ$ still hold?

Community
  • 1
yashirq
  • 505

1 Answers1

2

The definition must be rephrased as :

there are no truth assignment $A$ such that : $A(\psi)=1$ for every $\psi \in \Gamma$ and $A(\varphi)=0$.

Thus, what happens with a truth assignment $A'$ that assigns "false" (i.e. $0$) to some premise $\psi \in \Gamma$ ?

It doesn't matter what they "do" to $\varphi$, because we are only concerned with the truth assignements that "satisfy" (i.e. assign $1$) to all $\psi \in \Gamma$.

You must read the definition as a "recipe" to test for entailment :

(i) consider a truth assignement $A$ : are all formulae $\psi$ in $\Gamma$ true for $A$ ?

  • If NO, skip it (i.e. skip $A$).

  • If YES, then check if $\varphi$ is true.

(ii) if $\varphi$ is true for $A$, then take a new truth assignment $A'$ and go to step (i).

(iii) if $\varphi$ is false for $A$, then stop : $\varphi$ is not entailed by $\Gamma$.

Mauro ALLEGRANZA
  • 94,169
  • After rephrasing the definition, it means if $A(ψ)$=1 for every $ψ$∈$Γ$, then $A(φ)$ must be 1. Am I correct? – yashirq Oct 20 '16 at 06:42
  • I still don't get it. What happen if there exists A(ψ) =0 for ψ∈Γ? – yashirq Oct 20 '16 at 06:43
  • @yashirq - Comment 1 : CORRECT. – Mauro ALLEGRANZA Oct 20 '16 at 08:06
  • @yashirq - Comment 2 : following comment 1, if the antecedent of the conditional is false (because if for some $\psi \in \Gamma$ : $A'(\psi)=0$, we have that "$A'(\psi)=1$, for every $\psi \in \Gamma$" is false), then the conditional is true. This means that in this case the "statement" expressing the definition of entailment is satisfied. See vacuous truth. – Mauro ALLEGRANZA Oct 20 '16 at 08:09
  • Not quite get what you mean by saying the antecedent of the conditional is false then the conditional is true. I have some new understanding. If we have A$(ψ)$ =0, then the condition is false. But both 0 -> 0 and 0 -> 1 are true. So we don't care about what A$(φ)$ is. – yashirq Oct 20 '16 at 09:28
  • @yashirq - see Truth table for conditional. – Mauro ALLEGRANZA Oct 20 '16 at 09:35