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Proof step 148 from proof hasSharing-put-index-refval (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 147:
(st[r -- i ', refval(r2)][r2 -- sks3]) = refval(r1), st[r -- _type] .type = mkarraytype(ty), (st[r -- i ', refval(r2)][r2 -- (sks + i ' ' + sks0)]) = refval(r1), st[r2 -- sks] = refval(r), (st[r -- i ', refval(r2)][r2 -- sks]) = refval(r0), 0 < st[r -- _length] .intval, is_referencevalue(st[r -- i ', refval(r2)][r -- (i ' ' + sks + i ' ' + sks0)]), fits(r2, sks, st), fits(r0, i ' ' + sks0, st[r -- i ', refval(r2)]), i < st[r -- _length] .intval, 0 ≤ i, fits(r2, sks3, st[r -- i ', refval(r2)]), disjoint(str, r ' + r2 ', st), validref(str, r2, ty, st) ⊦ r = jvmref, r = r2, r1 = jvmref, r2 = jvmref, sks3 = [], sks + i ' ' + sks0 = sks3, junkref(str, r, st), hasSharing(r2, st), hasSharing(r, st), ∃ sks2. ¬ (sks2 + i ' ' ⊑ sks → st[r2 -- sks2] .refval ≠ r), ∃ sks1, sks4. sks = sks1 + i ' ' + sks4 ∧ st[r2 -- sks1] .refval = r ∧ (∀ sks2. sks2 + i ' ' ⊑ sks1 → st[r2 -- sks2] .refval ≠ st[r2 -- sks1] .refval)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, storekeys, Last, ⊦ sks + sk = sks + sk '], [CSIMPNAMEDSEQ : list, storekeys, c, ⊦ sk + sks = sk ' + sks], [CSIMPNAMEDSEQ : specialkeys, , type-03, ⊦ _type ≠ i '], [CSIMPNAMEDSEQ : specialkeys, , len, ⊦ _length ≠ i '])]]