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Proof step 4 from proof in-1init-06 (conclusion)

⊦ r ∈ addarray(r0, ty0, vals, st) ↔ r = r0 ∨ r ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ r ≠ r0 → (r ∈ addarray(r0, ty0, vals, st) ↔ r ∈ st)
Proof step 2:
⊦ r ∈ addarray(r, ty0, vals, st)
Proof step 3:
⊦ r ∈ st → r ∈ addarray(r0, ty0, vals, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]