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Proof step 271 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)), st[r4 -- sks] = refval(r0), st[r4 -- sks] = refval(r), (st[r -- i ', refval(r4)][r4 -- (sks + i ' ' + sks1)]) = refval(r1), st[r -- _type] .type = mkarraytype(ty), st[r -- sks0] = refval(r1), (st[r -- i ', refval(r4)][r0 -- i ']) = refval(r2), fits(r2, sks1, st[r -- i ', refval(r4)]), fits(r4, sks, st), disjoint(str, r ' + r4 ', st), validref(str, r4, ty, st), fits(r, sks0, st), i < st[r -- _length] .intval, 0 ≤ i, 0 < st[r -- _length] .intval, is_arraytype(st[r0 -- _type] .type) → i < st[r0 -- _length] .intval ⊦ r0 = jvmref, r4 = jvmref, sks0 = i ' ' + sks + i ' ' + sks1, r = jvmref, r = r4, r1 = jvmref, sks0 = [], is_classtype(st[r0 -- _type] .type), junkref(str, r, st), hasSharing(r4, st), hasSharing(r, st), ∃ sks3, sks4. sks = sks3 + i ' ' + sks4 ∧ st[r4 -- sks3] .refval = r ∧ (∀ sks2. sks2 + i ' ' ⊑ sks3 → st[r4 -- sks2] .refval ≠ st[r4 -- sks3] .refval), ∃ sks2. ¬ (sks2 + i ' ' ⊑ sks → st[r4 -- sks2] .refval ≠ r), ∃ sks2, sks1. sks0 = sks2 + i ' ' + sks1 ∧ st[r -- sks2] .refval = r ∧ (∀ sks3. sks3 + i ' ' ⊑ sks2 → st[r -- sks3] .refval ≠ r)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 270:
st[r0 -- sks0] = refval(r1), st[r0 -- _type] .type = mkarraytype(ty), (st[r0 -- i ', refval(r2)][r2 -- (sks + i ' ' + sks1)]) = refval(r1), st[r2 -- sks] = refval(r0), 0 < st[r0 -- _length] .intval, 0 ≤ i, i < st[r0 -- _length] .intval, fits(r0, sks0, st), validref(str, r2, ty, st), disjoint(str, r0 ' + r2 ', st), fits(r2, sks, st), fits(r2, sks1, st[r0 -- i ', refval(r2)]) ⊦ sks0 = [], r1 = jvmref, r0 = r2, r0 = jvmref, sks0 = i ' ' + sks + i ' ' + sks1, r2 = jvmref, hasSharing(r0, st), hasSharing(r2, st), junkref(str, r0, st), ∃ sks2, sks1. sks0 = sks2 + i ' ' + sks1 ∧ st[r0 -- sks2] .refval = r0 ∧ (∀ sks3. sks3 + i ' ' ⊑ sks2 → st[r0 -- sks3] .refval ≠ r0), ∃ sks2. ¬ (sks2 + i ' ' ⊑ sks → st[r2 -- sks2] .refval ≠ r0), ∃ sks3, sks4. sks = sks3 + i ' ' + sks4 ∧ st[r2 -- sks3] .refval = r0 ∧ (∀ sks2. sks2 + i ' ' ⊑ sks3 → st[r2 -- sks2] .refval ≠ st[r2 -- sks3] .refval)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : javatype, , test-16, ⊦ ¬ is_classtype (mkarraytype(ty))], [CSIMPNAMEDFORWARD : int-abs, , forward-le-05, ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPNAMEDSEQ : store, javastore, At-same, ⊦ (st[rk, val][rk]) = val], [CSIMPNAMEDSEQ : javavalue, , inj-03, ⊦ refval(r) = refval(r0) ↔ r = r0], [CSIMPNAMEDSEQ : list, storekeys, Last, ⊦ sks + sk = sks + sk '], [CSIMPNAMEDSEQ : list, storekeys, c, ⊦ sk + sks = sk ' + sks])]]