Solve Hamilton Circuit with Hamilton Path I want to show the reduction $HC \leq HP$.
Let $G=(V,E)$ be my undirected graph.

My idea is: For each edge $e=(u,v) \in E$ check whether $(V,E\backslash\{e\})$ has a Hamiltonian Path. If this is true for all edges, we have a Hamiltonian circuit in $G$. 

It is pretty trivial to proof the first direction (when we have a Hamiltonian circuit in $G$, we will always have a Hamiltonian path in $(V,E\backslash\{e\})$ - independent from the edge we choose).

I really have trouble finding a formal proof for the other direction (either


$\forall e\in E$ Hamiltonian path exists in $(V,E\backslash\{e\})$  $\Rightarrow$ Hamiltonian circle exists in $G$ or
There is no Hamiltonian circle in $E$ $\Rightarrow$ $\exists e \in E:$ $(V,E\backslash\{e\})$ has no Hamiltonian path
Does a graph have a hamiltonian cycle if for anyone of its edges, it has a hamiltonian path after the removal of that edge?

Note that any hypohamiltonian graph must be such a graph. Is it even true that every hypohamiltonian graph has a hamiltonian cycle?

Not necessarily.

A counterexample is the Peterson graph, which is hypohamiltonian but not hamiltonian.