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Proof step 623 from proof hasSharing-put-index-refval (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 622:
st[st[r2 -- sks3] .refval -- sks] = refval(r1), st[st[r2 -- sks3] .refval -- _type] .type = mkarraytype(ty), (st[st[r2 -- sks3] .refval -- i ', refval(r2)] [(st[st[r2 -- sks3] .refval -- i ', refval(r2)][r2 -- (sks3 + i ' ' + sks2)]) .refval -- (i ' ' + sks1) ] ) = refval(r1), st[r2 -- (sks3 + i ' ' + sks2)] = refval(st[r2 -- sks3] .refval), fits(st[r2 -- sks3] .refval, sks, st), disjoint(str, st[r2 -- sks3] .refval ' + r2 ', st), validref(str, r2, ty, st), i < st[st[r2 -- sks3] .refval -- _length] .intval, 0 ≤ i, fits(r2, sks3 + (i ' ' + sks2) + i ' ' + sks1, st[st[r2 -- sks3] .refval -- i ', refval(r2)]), 0 < st[st[r2 -- sks3] .refval -- _length] .intval, ∀ sks0. sks0 + i ' ' ⊑ sks3 → st[r2 -- sks0] .refval ≠ st[r2 -- sks3] .refval ⊦ st[r2 -- sks3] .refval = jvmref, st[r2 -- sks3] .refval = r2, r1 = jvmref, sks = [], r2 = jvmref, i ' ' + sks3 + (i ' ' + sks2) + i ' ' + sks1 = sks, junkref(str, st[r2 -- sks3] .refval, st), hasSharing(r2, st), hasSharing(st[r2 -- sks3] .refval, st), ∃ sks1, sks0. sks = sks1 + i ' ' + sks0 ∧ st[st[r2 -- sks3] .refval -- sks1] .refval = st[r2 -- sks3] .refval ∧ (∀ sks2. sks2 + i ' ' ⊑ sks1 → st[st[r2 -- sks3] .refval -- sks2] .refval ≠ st[st[r2 -- sks3] .refval -- sks1] .refval), ∃ sks0. ¬ (sks0 + i ' ' ⊑ sks3 + i ' ' + sks2 → st[r2 -- sks0] .refval ≠ st[r2 -- sks3] .refval)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDFORWARD : int-abs, , forward-le-05, ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPNAMEDSEQ : list, storekeys, Last, ⊦ sks + sk = sks + sk '], [CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPNAMEDSEQ : list, storekeys, e-03, ⊦ sks + sk ' + sks0 ≠ []], [CSIMPNAMEDSEQ : list, storekeys, c, ⊦ sk + sks = sk ' + sks])]]