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Proof step 25 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), fits(r, sks, st) ∧ fits(r, sks0, 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 3:
⊦ ¬ junkref(str, r, st0) ∧ fits(r, sks, st) ∧ st0 ⊆ st → fits(r, sks, st0)
Proof step 24:
¬ junkref(str, r, st) ∧ fits(r, sks0, st1) ∧ st ⊆ st1, 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), fits(r, sks, st) ∧ true ∧ st[r -- sks] .refval ≠ jvmref ∧ st[r -- sks] = st[r -- sks0]

Rule argument

[crewrite-arg: fits-subset , F , [str, r, st0, sks, st] → [str, r, st, sks0, st1] ]

Additional information

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