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Proof step 3 from proof in-init (conclusion)

⊦ r ≠ r0 → ((r -- sk) × val ∈ addarray(r0, ty, vals, st) ↔ (r -- sk) × val ∈ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ r ≠ r0 → (r -- sk ∈ addarray(r0, ty, vals, st) ↔ r -- sk ∈ st)
Proof step 2:
⊦ r ≠ r0 → (addarray(r, ty, vals, st)[r0 -- sk]) = st[r0 -- sk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk × val ∈ st ↔ ¬ ¬ (rk ∈ st ∧ st[rk] = val)], [CSIMPSEQ : ⊦ rk × val ∈ st ↔ rk ∈ st ∧ st[rk] = val])]]