bridge, n.

For e ∈ E(G),
e is a bridge iff
no cycle contains it.

Remove it,
and the graph
falls into two.

One crossing
lets you walk it.

Certifying it
takes the whole graph.

Call this frame F.

e ∈ E(F)

No return path exists.
No return path
appears.

Whether e ∈ B(G)
depends on G.

F = G

F is not G.

Past where the leaves close over it,
the path could double back,
open a second way
to the same two sides,
and un-bridge this exact plank
without moving it
an inch.

e ∈ B(G) is not undefined.

It is unproven,
from here.