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Proof step 6 from proof is_obj-addarray (previous step)

⊦ r = r0, class+∃(str, (addarray(r0, ty0, vals, st)[r -- _type]) .type .class) ∧ ¬ is_abstract((addarray(r0, ty0, vals, st)[r -- _type]) .type .class) ∧ is_obj(r, addarray(r0, ty0, vals, st)) ↔ is_obj(str, r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ r ≠ r0 → (addarray(r, ty, vals, st)[r0 -- sk]) = st[r0 -- sk]
Proof step 3:
⊦ r1 ≠ r0 → (is_obj(r1, addarray(r0, ty, vals, st)) ↔ is_obj(r1, st))
Proof step 5:
⊦ r = r0, class+∃(str, st[r -- _type] .type .class) ∧ ¬ is_abstract(st[r -- _type] .type .class) ∧ is_obj(r, st) ↔ is_obj(str, r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]