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Proof step 256 from proof hasSharing-put-index-refval (previous step)

¬ ( ¬ (∃ sks0, sks1. sks = sks0 + (i ' + sks1) ∧ st[r4 -- sks0] .refval = r ∧ (∀ sks2. sks2 + i ' ⊑ sks0 → st[r4 -- sks2] .refval ≠ r)) ∧ i ' ≠ _type ∧ i ' ≠ _length), st[r -- sks0] = refval(r1), st[r -- _type] .type = mkarraytype(ty), (st[r -- i ', refval(r4)][r4 -- (sks + i ' ' + sks1)]) = refval(r1), st[r4 -- sks] = refval(r), (st[r -- i ', refval(r4)][r4 -- sks]) = refval(r0), 0 < st[r -- _length] .intval, 0 ≤ i, i < st[r -- _length] .intval, fits(r, sks0, st), validref(str, r4, ty, st), disjoint(str, r ' + r4 ', st), fits(r4, sks, st[r -- i ', refval(r4)]), fits(r0, i ' ' + sks1, st[r -- i ', refval(r4)]) ⊦ sks0 = [], r1 = jvmref, r = r4, r = jvmref, sks0 = i ' ' + sks + i ' ' + sks1, r4 = jvmref, hasSharing(r, st), hasSharing(r4, st), junkref(str, r, st), ∃ sks2, sks1. sks0 = sks2 + i ' ' + sks1 ∧ st[r -- sks2] .refval = r ∧ (∀ sks3. sks3 + i ' ' ⊑ sks2 → st[r -- sks3] .refval ≠ r), ∃ sks2. ¬ (sks2 + i ' ' ⊑ sks → st[r4 -- sks2] .refval ≠ r)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 255:
(st[st[r4 -- sks3] .refval -- i ', refval(r4)][r4 -- (sks3 + i ' ' + sks4)]) = refval(r0), st[r4 -- (sks3 + i ' ' + sks4)] = refval(st[r4 -- sks3] .refval), (st[st[r4 -- sks3] .refval -- i ', refval(r4)][r4 -- ((sks3 + i ' ' + sks4) + i ' ' + sks1)]) = refval(r1), st[st[r4 -- sks3] .refval -- _type] .type = mkarraytype(ty), st[st[r4 -- sks3] .refval -- sks0] = refval(r1), fits(r0, i ' ' + sks1, st[st[r4 -- sks3] .refval -- i ', refval(r4)]), fits(r4, sks3 + i ' ' + sks4, st[st[r4 -- sks3] .refval -- i ', refval(r4)]), disjoint(str, st[r4 -- sks3] .refval ' + r4 ', st), validref(str, r4, ty, st), fits(st[r4 -- sks3] .refval, sks0, st), i < st[st[r4 -- sks3] .refval -- _length] .intval, 0 ≤ i, 0 < st[st[r4 -- sks3] .refval -- _length] .intval ⊦ r4 = jvmref, sks0 = i ' ' + (sks3 + i ' ' + sks4) + i ' ' + sks1, st[r4 -- sks3] .refval = jvmref, st[r4 -- sks3] .refval = r4, r1 = jvmref, sks0 = [], junkref(str, st[r4 -- sks3] .refval, st), hasSharing(r4, st), hasSharing(st[r4 -- sks3] .refval, st), ∃ sks2. ¬ (sks2 + i ' ' ⊑ sks3 → st[r4 -- sks2] .refval ≠ st[r4 -- sks3] .refval), ∃ sks2. ¬ (sks2 + i ' ' ⊑ sks3 + i ' ' + sks4 → st[r4 -- sks2] .refval ≠ st[r4 -- sks3] .refval), ∃ sks2, sks1. sks0 = sks2 + i ' ' + sks1 ∧ st[st[r4 -- sks3] .refval -- sks2] .refval = st[r4 -- sks3] .refval ∧ (∀ sks. sks + i ' ' ⊑ sks2 → st[st[r4 -- sks3] .refval -- sks] .refval ≠ st[r4 -- sks3] .refval)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDFORWARD : int-abs, , forward-le-05, ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPNAMEDSEQ : specialkeys, , len, ⊦ _length ≠ i '], [CSIMPNAMEDSEQ : specialkeys, , type-03, ⊦ _type ≠ i '], [CSIMPNAMEDSEQ : list, storekeys, Last, ⊦ sks + sk = sks + sk '], [CSIMPNAMEDSEQ : list, storekeys, c, ⊦ sk + sks = sk ' + sks])]]