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Proof step 10 from proof hasSharing-subset (previous step)

st1[r -- sks0] = refval(r0), st1[r -- sks] = refval(r0), fits(r, sks, st1), fits(r, sks0, st1), st ⊆ st1 ⊦ sks0 = sks, sks0 = [], sks = [], r0 = jvmref, junkref(str, r, st), r -- sks ∈ st, st[r -- sks] .refval ≠ jvmref ∧ st[r -- sks] = st[r -- sks0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ sks ≠ [] ∧ ¬ junkref(str, r, st) ∧ fits(r, sks, st) → r -- sks ∈ st
Proof step 9:
¬ (sks ≠ [] ∧ ¬ junkref(str, r, st) ∧ fits(r, sks, st)), st1[r -- sks0] = refval(r0), st1[r -- sks] = refval(r0), fits(r, sks, st1), fits(r, sks0, st1), st ⊆ st1 ⊦ sks0 = sks, sks0 = [], sks = [], r0 = jvmref, junkref(str, r, st), r -- sks ∈ st, st[r -- sks] .refval ≠ jvmref ∧ st[r -- sks] = st[r -- sks0]

Rule argument

[crewrite-arg: fits-in, F, [sks, str, r, st] → [sks, str, r, st]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sks, str, r, st)]]