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Proof step 19 from proof hasSharing-subset (previous step)

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)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ st ⊆ st0 ∧ r -- sks ∈ st ∧ sks ≠ [] → st0[r -- sks] = st[r -- sks]
Proof step 18:
¬ (st ⊆ st1 ∧ r -- sks0 ∈ st ∧ sks0 ≠ []), 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

[speclemmaarg: ,st ⊆ st0 ∧ r -- sks ∈ st ∧ sks ≠ [] → st0[r -- sks] = st[r -- sks] ,[st, st0, r, sks] → [st, st1, r, sks0] ,F ,storefuns , ,lookup-subset ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, st1, r, sks0)]]