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Proof step 31 from proof hasSharing-subset (previous step)

st ⊆ st1, hasSharing(r, st1) ⊦ junkref(str, r, st), ∃ sks1, sks2. fits(r, sks1, st) ∧ fits(r, sks2, st) ∧ sks1 ≠ sks2 ∧ sks1 ≠ [] ∧ sks2 ≠ [] ∧ st[r -- sks1] .refval ≠ jvmref ∧ st[r -- sks1] = st[r -- sks2]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ hasSharing(r, st) ↔ (∃ sks1, sks2. fits(r, sks1, st) ∧ fits(r, sks2, st) ∧ sks1 ≠ sks2 ∧ sks1 ≠ [] ∧ sks2 ≠ [] ∧ st[r -- sks1] .refval ≠ jvmref ∧ st[r -- sks1] = st[r -- sks2])
Proof step 30:
st ⊆ st1, ∃ sks1, sks2. fits(r, sks1, st1) ∧ fits(r, sks2, st1) ∧ sks1 ≠ sks2 ∧ sks1 ≠ [] ∧ sks2 ≠ [] ∧ st1[r -- sks1] .refval ≠ jvmref ∧ st1[r -- sks1] = st1[r -- sks2] ⊦ junkref(str, r, st), ∃ sks1, sks2. fits(r, sks1, st) ∧ fits(r, sks2, st) ∧ sks1 ≠ sks2 ∧ sks1 ≠ [] ∧ sks2 ≠ [] ∧ st[r -- sks1] .refval ≠ jvmref ∧ st[r -- sks1] = st[r -- sks2]

Rule argument

[crewrite-arg: hasSharing-def, F, [st, r] → [st1, r]]

Additional information

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