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Proof step 75 from proof hasSharing-put-index-refval (previous step)

sks ≠ [] ∧ i ' ' + sks1 ≠ [], (st[r0 -- i ', refval(r1)][r0 -- sks]) = refval(r), st[r0 -- sks] = refval(r0), (st[r0 -- i ', refval(r1)][(st[r0 -- i ', refval(r1)][r0 -- sks]) .refval -- (i ' ' + sks1)]) = refval(r1), st[r0 -- _type] .type = mkarraytype(ty), (st[r0 -- i ', refval(r1)][r -- i ']) = refval(r2), fits(r2, sks1, st[r0 -- i ', refval(r1)]), fits(r0, sks, st), validref(str, r1, ty, st), disjoint(str, r0 ' + r1 ', st), 0 < st[r0 -- _length] .intval, 0 ≤ i, i < st[r0 -- _length] .intval, is_arraytype(st[r -- _type] .type) → i < st[r -- _length] .intval ⊦ r = jvmref, r1 = jvmref, r0 = r1, r0 = jvmref, sks + i ' ' + sks1 = i ' ', is_classtype(st[r -- _type] .type), hasSharing(r0, st), hasSharing(r1, st), junkref(str, r0, st), ∃ sks2. ¬ (sks2 + i ' ' ⊑ sks → st[r0 -- sks2] .refval ≠ r0), ∃ sks0, sks3. sks = sks0 + i ' ' + sks3 ∧ st[r0 -- sks0] .refval = r0 ∧ (∀ sks2. sks2 + i ' ' ⊑ sks0 → st[r0 -- sks2] .refval ≠ r0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPNAMEDSEQ : list, storekeys, e-01, ⊦ sk ' + sks ≠ []])]]