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The call from DDD to H0(DDD) when N steps of DDD are correctlyYes…
emulated by any pure function x86 emulator H0 cannot possibly
return.
The behavior of the directly executed DDD() is irrelevantDDD is literally the input. How can you say that is not the
because that is not the behavior of the input. Deciders
compute the mapping from their actual finite string input
to an output by a sequence of finite string transformations.
In this case the sequence is the line-by-line executionThe behaviour of DDD is not determined by the H0 simulating
trace of the behavior of DDD correctly emulated by H0.
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