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Proof step 24 from proof hasSharing-def-better (previous step)

st[r -- sks] = refval(r0), st[r -- sks0] = refval(r0), fits(r, sks0, st), fits(r, sks, st) ⊦ r0 = jvmref, sks0 = sks, ∃ 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], sks0 ≠ [] ∧ sks ≠ []

User applied rule cut formula

And obtained the following new goals

Proof step 14:
sks = [], st[r -- sks] = refval(r0), st[r -- sks0] = refval(r0), fits(r, sks0, st), fits(r, sks, st) ⊦ r0 = jvmref, sks0 = sks, ∃ 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], sks0 ≠ [] ∧ sks ≠ []
Proof step 23:
st[r -- sks] = refval(r0), st[r -- sks0] = refval(r0), fits(r, sks0, st), fits(r, sks, st) ⊦ sks = [], r0 = jvmref, sks0 = sks, ∃ 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], sks0 ≠ [] ∧ sks ≠ []

Rule argument

[fma-arg: sks = []]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (sks = [])]]