Re: [isabelle] Problem with arbitrary specification in double inductive rule induction



Hi Mark,

Why is m1 still free even though I have specified that it should be
arbitrary? Including m1 in the arbitrary specification makes no difference however removing m from the specification does make m free (as expected).

try this:

proof(induction arbitrary: m and m1 rule: P_Q.inducts)

Does that yield the proof state that you expected?

Cheers
Lars




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