[isabelle] Transfer rule for undefined



Dear developers of lifting/transfer,

First of all, I'd like to thank all of you for this great tool, I am now using it all the time. Unfortunately, I keep proving different transfer rules for the constant undefined over and over again, although they all have the same shape.

The following example illustrates my setting:

typedef my_int = "UNIV :: int set" .. setup_lifting type_definition_my_int
typedef my_nat = "UNIV :: nat set" .. setup_lifting type_definition_my_nat

lift_definition P :: "my_int => bool" is "op > 0" .
lift_definition foo :: "my_int => bool => my_nat" is "%i _. nat i" .
lift_definition my_int_of_my_nat :: "my_nat => my_int" is int .

definition bar :: "my_int => bool => my_nat"
  where "bar i b = (if P i then undefined i b else foo i b)"

lemma "foo (my_int_of_my_nat n) b = bar (my_int_of_my_nat n) b"
  unfolding bar_def
  apply transfer

This gives me the following second subgoal for undefined:

  Transfer.Rel (fun_rel cr_my_int (fun_rel op = cr_my_nat)) ?ah23 undefined

So far, I just proved this transfer rule for an appropriate instantiation of ?ah23, but I have to prove similar goals with different combinations of fun_rel, cr_... etc. So I tried to prove a generic transfer lemma for quotients:

lemma undefined_transfer:
  assumes Q1: "Quotient A Abs1 Rep1 cr1"
  and Q2: "Quotient B Abs2 Rep2 cr2"
  shows "(fun_rel cr2 cr1) (Rep1 o undefined o Abs2) undefined"
by(auto dest!: Q2[unfolded Quotient_alt_def, THEN conjunct1, rule_format]
     intro!: Q1[unfolded Quotient_alt_def, THEN conjunct2, THEN conjunct1, rule_format])

With this lemma, I can prove all these rules for undefined -- in the running example:

apply(unfold Rel_def)
apply(rule undefined_transfer fun_quotient identity_quotient
        Quotient_my_nat Quotient_my_int)+

Unfortunately, I did not manage to have transfer prove these rules on the fly. How can I get there? The following declarations do not suffice:

lemmas [transfer_rule] =
  undefined_transfer Quotient_my_int Quotient_my_nat fun_quotient identity_quotient

Thanks in advance for any help,
Andreas

PS: The example is from Isabelle2013, but it is similar in the development version (id 13171b27eaca).




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