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Proof step 4 from proof in-1dim-04 (previous step)

∃ sk. r -- sk ∈ addarray(r0, val, i, st) ⊦ r = r0, r ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ r ≠ r0 → (r -- sk ∈ addarray(r0, val, i, st) ↔ r -- sk ∈ st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st])]]