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Proof step 589 from proof hasSharing-put-index-refval (previous step)

¬ ( ¬ (∃ sks0, sks2. sks1 = sks0 + (i ' + sks2) ∧ st[r -- sks0] .refval = r0 ∧ (∀ sks2. sks2 + i ' ⊑ sks0 → st[r -- sks2] .refval ≠ r0)) ∧ i ' ≠ _type ∧ i ' ≠ _length), (st[r0 -- i ', refval(r2)][r -- (sks1 + i ' ' + sks2)]) = refval(r1), st[r0 -- _type] .type = mkarraytype(ty), st[r -- sks] = refval(r1), st[r -- sks1] = refval(r0), (st[r0 -- i ', refval(r2)][r -- sks1]) = refval(r3), validref(str, r2, ty, st), disjoint(str, r ' + r2 ', st), fits(r, sks, st), fits(r, sks1, st[r0 -- i ', refval(r2)]), fits(r3, i ' ' + sks2, st[r0 -- i ', refval(r2)]), r2 = jvmref → is_reftype(ty) ⊦ sks1 + i ' ' + sks2 = sks, sks = [], r1 = jvmref, r = r2, r0 = jvmref, hasSharing(r, st), hasSharing(r2, st), junkref(str, r, st), ∃ sks0, sks3. sks = sks0 + i ' ' + sks3 ∧ st[r -- sks0] .refval = r0 ∧ (∀ sks2. sks2 + i ' ' ⊑ sks0 → st[r -- sks2] .refval ≠ st[r -- sks0] .refval), ∃ sks3. ¬ (sks3 + i ' ' ⊑ sks1 → st[r -- sks3] .refval ≠ r0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 588:
(st[st[r -- sks0] .refval -- i ', refval(r2)][r -- (sks0 + i ' ' + sks3)]) = refval(r3), st[r -- (sks0 + i ' ' + sks3)] = refval(st[r -- sks0] .refval), st[r -- sks] = refval(r1), st[st[r -- sks0] .refval -- _type] .type = mkarraytype(ty), (st[st[r -- sks0] .refval -- i ', refval(r2)][r -- ((sks0 + i ' ' + sks3) + i ' ' + sks2)]) = refval(r1), fits(r3, i ' ' + sks2, st[st[r -- sks0] .refval -- i ', refval(r2)]), fits(r, sks0 + i ' ' + sks3, st[st[r -- sks0] .refval -- i ', refval(r2)]), fits(r, sks, st), disjoint(str, r ' + r2 ', st), validref(str, r2, ty, st), r2 = jvmref → is_reftype(ty) ⊦ st[r -- sks0] .refval = jvmref, r = r2, r1 = jvmref, sks = [], (sks0 + i ' ' + sks3) + i ' ' + sks2 = sks, junkref(str, r, st), hasSharing(r2, st), hasSharing(r, st), ∃ sks4. ¬ (sks4 + i ' ' ⊑ sks0 → st[r -- sks4] .refval ≠ st[r -- sks0] .refval), ∃ sks. ¬ (sks + i ' ' ⊑ sks0 + i ' ' + sks3 → st[r -- sks] .refval ≠ st[r -- sks0] .refval), ∃ sks1, sks3. sks = sks1 + i ' ' + sks3 ∧ st[r -- sks1] .refval = st[r -- sks0] .refval ∧ (∀ sks2. sks2 + i ' ' ⊑ sks1 → st[r -- sks2] .refval ≠ st[r -- sks1] .refval)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [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])]]