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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ is_valid_ref(r, st) ↔ is_obj(r, st) ∨ is_array(r, st)
Proof step 2:
⊦ r ≠ r0 → (is_obj(r, addobj(r0, class, fis, st)) ↔ is_obj(r, st))
Proof step 3:
⊦ r ≠ r0 → (is_array(r, addobj(r0, class, fis, st)) ↔ is_array(r, st))
Proof step 4:
⊦ ¬ is_obj(r, st) → (is_valid_ref(r, st) ↔ is_array(r, st))
Proof step 5:
⊦ is_obj(r, st) → is_valid_ref(r, st)
Proof step 6:
⊦ is_array(r, st) → is_valid_ref(r, st)

Rule argument

[crewrite-arg: is-valid-ref , F , [st, r] → [addobj(r0, class, fis, st), r] ]

Additional information

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