$L=\{\left<M\right> \ | \ M $ is a TM s.t. $M$ does not accept any string starting with a '1' $\}$. Assume the alphabet to be $\Sigma = \{0,1\}$.
By Rice's theorem $L$ is undecidable. I managed to "prove" that $L$ is decidable. So can someone point out the error in my proof. To keep things clean I ignore the empty string since it can easily be dealt with as a special case.
We can write $$L = \{\left<M\right> \ | \ w\in \Sigma^*, \ w \in L(M) \implies w[0] \neq 1\}$$ $$ = \{\left<M\right> \ | \ w\in \Sigma^*, \ w \in L(M) \implies w[0] = 0 \}$$ To get $\overline{L}$ we need to negate the condition in $L$. Hence $$\overline{L}= \{\left<M\right> \ | \ w\in \Sigma^*, \ w \in L(M) \implies w[0] = 1 \}$$
We have $$\exists \ w\in \Sigma^* \text{ such that } w\in L(M) \text{ and } w[0] = 1 \iff \left<M\right>\not \in L \iff \left<M\right> \in \overline{L}$$ We can construct a recognizer for $\overline{L}$ by searching for such a $w$
$R=$ 'on input $\left<M\right>$ where $M$ is a valid TM:
for $w \in \Sigma^*$:
$\ \ $if $M(w) =$ accept
$\ \ \ \ $if $w[0]=1$ then accept'
So $\overline{L}$ is recognizable. Similarly we have
$$\exists \ w\in \Sigma^* \text{ such that } w\in L(M) \text{ and } w[0] = 0 \iff \left<M\right>\not \in \overline{L} \iff \left<M\right> \in L$$ We can construct a recognizer for $L$ by searching for such a $w$
$R'=$ 'on input $\left<M\right>$ where $M$ is a valid TM:
for $w \in \Sigma^*$:
$\ \ $if $M(w) =$ accept
$\ \ \ \ $if $w[0]=0$ then accept'
So $L$ is also recognizable, this means $L$ is decidable.