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Proof step 7 from proof is_array-addobj (previous step)

⊦ r = r0, is_array(str, r, addobj(r0, class, fis, st)) ↔ is_array(str, r, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ is_array(str, r, st) ↔ type∃(str, st[r -- _type] .type) ∧ st[r -- _type] .type .jtelemtype ≠ void_type ∧ st[r -- _type] .type .jtelemtype ≠ abstract_type ∧ is_array(r, st)
Proof step 6:
⊦ r = r0, type∃(str, (addobj(r0, class, fis, st)[r -- _type]) .type) ∧ (addobj(r0, class, fis, st)[r -- _type]) .type .jtelemtype ≠ void_type ∧ (addobj(r0, class, fis, st)[r -- _type]) .type .jtelemtype ≠ abstract_type ∧ is_array(r, addobj(r0, class, fis, st)) ↔ is_array(str, r, st)

Rule argument

[speclemmaarg: , is_array(str, r, st) ↔ type∃(str, st[r -- _type] .type) ∧ st[r -- _type] .type .jtelemtype ≠ void_type ∧ st[r -- _type] .type .jtelemtype ≠ abstract_type ∧ is_array(r, st) ,[st, r, str] → [addobj(r0, class, fis, st), r, str] ,F ,okfuns , ,is-real-array-def ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (addobj(r0, class, fis, st), r, str)]]