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Proof step 15 from proof hasSharing-array-forward (previous step)

st[r0 -- sks0] = refval(r1), st[r -- i '] = refval(r0), st[r0 -- sks] = refval(r1), 0 < st[r -- _length] .intval, 0 ≤ i, is_arraytype(st[r -- _type] .type), i < st[r -- _length] .intval, fits(r0, sks, st), fits(r0, sks0, st) ⊦ sks0 = sks, sks0 = [], sks = [], r1 = jvmref, r = jvmref, ∃ 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 exists right

And obtained the following new goals

Proof step 14:
st[r0 -- sks0] = refval(r1), st[r -- i '] = refval(r0), st[r0 -- sks] = refval(r1), 0 < st[r -- _length] .intval, 0 ≤ i, is_arraytype(st[r -- _type] .type), i < st[r -- _length] .intval, fits(r0, sks, st), fits(r0, sks0, st) ⊦ sks0 = sks, sks0 = [], sks = [], r1 = jvmref, r = jvmref, fits(r, i ' + sks0, st) ∧ fits(r, i ' + sks, st) ∧ i ' + sks0 ≠ i ' + sks ∧ i ' + sks0 ≠ [] ∧ i ' + sks ≠ [] ∧ st[r -- (i ' + sks0)] .refval ≠ jvmref ∧ st[r -- (i ' + sks0)] = st[r -- (i ' + sks)]

Rule argument

[EXRARG : pos: right 6, [extquanttermlist: (i ' + sks0, i ' + sks) abort:F , discard:T , computed:F ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (i ' + sks0, i ' + sks)]]