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Proof step 7 from proof stsubset-new (previous step)

st[rk] = ((rk = r -- sk ⊃ val; st0[rk])), rk ∈ st, rk ∈ st0[r -- sk, val], ∀ rk. rk ∈ st → st[rk] = ((rk = r -- sk ⊃ val; st0[rk])) ∧ rk ∈ st0[r -- sk, val], ∀ rk. rk ∈ st → rk .ref ≠ r ⊦ r = jvmref, rk ∈ st0 ∧ st[rk] = st0[rk]

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 6:
rk = r -- sk, st[rk] = val, rk ∈ st, rk ∈ st0[r -- sk, val], ∀ rk. rk ∈ st → st[rk] = ((rk = r -- sk ⊃ val; st0[rk])) ∧ rk ∈ st0[r -- sk, val], ∀ rk. rk ∈ st → rk .ref ≠ r ⊦ r = jvmref, rk ∈ st0 ∧ st[rk] = st0[rk]

Rule argument

[CASEDARG : pos: left 1, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st])]]