Assumptions: groupoid, one-way
Conclusions: thin
Reason: If f,g:A⇉Bf,g : A \rightrightarrows Bf,g:A⇉B are two morphisms, then f−1∘g:A→Af^{-1} \circ g : A \to Af−1∘g:A→A must be the identity, so that f=gf = gf=g.