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Proof step 9 from proof stsubset-03 (previous step)

rk .ref ≠ r, st[rk] = st0[rk], newref(r, st), rk ∈ st, rk ∈ addarray(r, ty, i, st0), ∀ rk. rk ∈ st → rk ∈ addarray(r, ty, i, st0) ∧ st[rk] = (addarray(r, ty, i, st0)[rk]) ⊦ r = jvmref, rk ∈ st0 ∧ st[rk] = st0[rk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
st[rk] = noval, newref(r, st), rk ∈ st, rk ∈ addarray(r, ty, i, st0), ∀ rk. rk ∈ st → rk ∈ addarray(r, ty, i, st0) ∧ st[rk] = (addarray(r, ty, i, st0)[rk]) ⊦ r = jvmref, rk .ref = r, rk ∈ st0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ∅[rk] = noval], [CSIMPSEQ : ⊦ ¬ rk ∈ st → st[rk] = ∅[rk]])]]