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Proof step 21 from proof same-subclasses-is-array (previous step)

has_classtype(st[r -- _type] .type), type∃(str0, st[r -- _type] .type), same-subclasses(st[r -- _type] .type .jtelemtype .class, str, str0), is_array(r, st) ⊦ st[r -- _type] .type .jtelemtype = abstract_type, st[r -- _type] .type .jtelemtype = void_type, tdname+∃(str, st[r -- _type] .type .jtelemtype .class)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ type∃(str, ty) ↔ ¬ has_classtype(ty) ∨ tdname+∃(str, ty .jtclass)
Proof step 20:
has_classtype(st[r -- _type] .type), ¬ has_classtype(st[r -- _type] .type) ∨ tdname+∃(str0, st[r -- _type] .type .jtclass), same-subclasses(st[r -- _type] .type .jtelemtype .class, str, str0), is_array(r, st) ⊦ st[r -- _type] .type .jtelemtype = abstract_type, st[r -- _type] .type .jtelemtype = void_type, tdname+∃(str, st[r -- _type] .type .jtelemtype .class)

Rule argument

[speclemmaarg: ,type∃(str, ty) ↔ ¬ has_classtype(ty) ∨ tdname+∃(str, ty .jtclass) ,[ty, str] → [st[r -- _type] .type, str0] ,F ,typefuns , ,typeex-def ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st[r -- _type] .type, str0)]]