Will current flow beyond the diode?
Let me redraw the central part of the circuit. Without the diode being present, current will flow clockwise. When D1 is inserted, it is reverse-biased, because V2 is higher than V1 (they are in parallel). Reverse-biased means no current flows.

From then on, you can treat this entire thing as its Thevenin equivalent - no matter what you do the path V1-D1 will always be inert, because V2 makes sure D1 stays reverse-biased.







simulate this circuit &ndash; Schematic created using CircuitLab



Just to clarify this issue, let's make a thought experiment.


The circuit of V1, V2, D1 is clearly a sub-circuit of the circuit in question.
Voltage between NODE1 and NODE2 is guaranteed to be 20V, since V2 is connected directly between the two. (V2 being an ideal voltage source, voltage across it cannot change by definition.)
Since V1-D1 and V2 are in parallel, they share the same voltage across them.
When you subtract 5V from 20V (this is perfectly legal with voltage sources), you're left with 15V between D1's anode and cathode.
Diode will act as an open circuit, since the potential at its cathode is higher than its anode = it is reverse biased.