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

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

User applied rule all left

And obtained the following new goals

Proof step 8:
rk ∈ st, ∀ rk. rk ∈ st → rk .ref ≠ r, rk ∈ st → st[rk] = ((rk = r -- sk ⊃ val; st0[rk])) ∧ rk ∈ st0[r -- sk, val], ∀ rk. rk ∈ st → st[rk] = ((rk = r -- sk ⊃ val; st0[rk])) ∧ rk ∈ st0[r -- sk, val] ⊦ r = jvmref, rk ∈ st0 ∧ st[rk] = st0[rk]

Rule argument

[EXRARG : pos: left 3, [extquanttermlist: (rk)abort:F, discard:F, computed:T]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rk)]]