Editing Peterson's algorithm (section)

=== Mutual exclusion ===
{{Main|Mutual exclusion}}
P0 and P1 can never be in the critical section at the same time. If P0 is in its critical section, then <code>flag[0]</code> is true. In addition, either <code>flag[1]</code> is <code>false</code> (meaning that P1 has left its critical section), or <code>turn</code> is <code>0</code> (meaning that P1 is just now trying to enter the critical section, but graciously waiting), or P1 is at label <code>P1_gate</code> (trying to enter its critical section, after setting <code>flag[1]</code> to <code>true</code> but before setting <code>turn</code> to <code>0</code> and busy waiting). So if both processes are in their critical sections, then we conclude that the state must satisfy <code>flag[0]</code> and <code>flag[1]</code> and <code>turn = 0</code> and <code>turn = 1</code>. No state can satisfy both <code>turn = 0</code> and <code>turn = 1</code>, so there can be no state where both processes are in their critical sections.
(This recounts an argument that is made rigorous in Schneider 1997.<ref name="Schneider.p185">F. B. Schneider, ''On Concurrent Programming'', Springer Verlag, 1997, pages 185–196.</ref>)