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Proof step 20 from proof hasSharing-subset (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
st[r -- sks] = refval(r0), st1[r -- sks0] = refval(r0), st ⊆ st1, fits(r, sks0, st1), fits(r, sks, st1), r -- sks ∈ st ⊦ sks = [], r0 = jvmref, sks0 = [], sks0 = sks, refval(r0) = st[r -- sks0], junkref(str, r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ refval(r) .refval = r])]]