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Proof step 6 from proof in-1dim-04 (conclusion)

⊦ r ≠ r0 → (r ∈ addarray(r0, val, i, st) ↔ r ∈ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ r ∈ st → r ∈ addarray(r0, val, i, st)
Proof step 5:
r ∈ addarray(r0, val, i, st) ⊦ r = r0, r ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]