If $s \to t$ is satisfiable and $\neg s \to t$ is satisfiable then $t$ is satisfiable

Question

I am struggling with the following question:

Is the following statement true?

If $s \to t$ is satisfiable and $\neg s \to t$ is satisfiable then $t$ is satisfiable.

I thought this was true, but my answer sheet says "false" with no explanation. Can anyone see why?

Answer 1

Don't forget that an implication is always true if the hypothesis is wrong. In particular, $\bot\implies \bot$ is true, even though $\bot$ is never true by definition.

So, for example, both $s\implies \bot$ and $\lnot s\implies \bot$ have models, i.e. the left one where $s$ is false, and the right one where $s$ is true.
Yet, still, $\bot$ remains false in every model.

Answer 2

Satisfiable is the not the same as always true.

If $s\to t$ and $\lnot s\to t$ were always true, then $t$ would be always true.

But to say $s\to t$ and $\lnot s \to t$ are satisfiable only says that $s\to t$ and $\lnot s\to t$ are each sometimes true, which is not a strong enough condition to force $t$ to be sometimes true.

For example, let $s=p$, where $p$ is a boolean variable, and let $t=\bot$ (or any proposition that is always false).

If $p$ is set to false, then $s\to t$ is true, and if $p$ is set to true, then $\lnot s\to t$ is true, hence both $s\to t$ and $\lnot s\to t$ are satisfiable, but $t$ is not satisfiable.

Answer 3

Satisfiable means that :

it is possible to find an interpretation (model) that makes the formula true.

In propositional logic, this amounts to find a suitable truth assignment $v$ such that the formula is evaluated to TRUE by $v$.

Consider a truth assignment $v$ such that $v(s)=v(t)=$ FALSE; we have that $v(s \to t)=$ TRUE, and thus $s \to t$ is satisfiable.

Consider now a truth assignment $v'$ such that $v(s)=$ TRUE and $v(t)=$ FALSE; we have that $v(\lnot s \to t)=$ TRUE, and thus also $\lnot s \to t$ is satisfiable.

But in both cases $t$ is evaluated to FALSE.