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Proof step 9 from proof in-init-02 (conclusion)

⊦ r ≠ r0 → (r -- sk ∈ addarray(r0, vals, j, st) ↔ r -- sk ∈ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ rk ∈ st → rk ∈ addarray(r, vals, j, st)
Proof step 2:
⊦ r ≠ r0 → (r ∈ addarray(r0, vals, j, st) ↔ r ∈ st)
Proof step 8:
r -- sk ∈ addarray(r0, vals, j, st), r ∈ st ⊦ r = r0, r -- sk ∈ st

Rule argument

[emptyarg]

Additional information

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