Pipe-calculus: Difference between revisions

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Any other combination of operands leads to failure of the current fork.
Any other combination of operands leads to failure of the current fork.
The following example fails because synchronisation cannot occur between different atoms.


<math>a^+ . s \rhd b^- . t = \mathsf{fail}</math>
<math>a^+ . s \rhd b^- . t = \mathsf{fail}</math>
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