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Proof step 10 from proof stsubset-03 (previous step)

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]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ rk .ref ≠ r → (addarray(r, ty, i, st)[rk]) = st[rk]
Proof step 6:
¬ 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]
Proof step 9:
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]

Rule argument

[crewrite-arg: add-at-02, F, [rk, st, i, ty, r] → [rk, st0, i, ty, r]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rk, st0, i, ty, r)]]