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Proof step 6 from proof stsubset-03 (previous step)

¬ rk .ref ≠ r, st[rk] = (addarray(r, ty, i, 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 5:
st[rk] = (addarray(rk .ref, ty, i, st0)[rk]), newref(rk .ref, st), rk ∈ st, rk ∈ addarray(rk .ref, ty, i, st0), ∀ rk0. rk0 ∈ st → rk0 ∈ addarray(rk .ref, ty, i, st0) ∧ st[rk0] = (addarray(rk .ref, ty, i, st0)[rk0]) ⊦ rk .ref = jvmref, rk ∈ st0 ∧ st[rk] = st0[rk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ newref(r, st) → r ≠ jvmref])]]