1. (inl $\cdot$ ) = (inr $\cdot$ )
\\[0ex]2. case inl $\cdot$  of inl($x$) =$>$ 0 $\mid$ inr($x$) =$>$ 1 = case inr $\cdot$  of inl($x$) =$>$ 0 $\mid$ inr($x$) =$>$ 1
\\[0ex]$\vdash$  False