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Proof step 29 from proof isobj-addarray (conclusion)

⊦ r1 ≠ r0 → (is_obj(r1, addarray(r0, ty, vals, st)) ↔ is_obj(r1, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ r ≠ r0 ∧ is_obj(r, st) → is_obj(r, addarray(r0, ty0, vals, st))
Proof step 28:
is_obj(r1, addarray(r0, ty, vals, st)) ⊦ r1 = r0, is_obj(r1, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]