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Proof step 273 from proof hasSharing-put-index-refval (previous step)

(st[r -- i ', refval(r4)][r0 -- i ']) = refval(r2), 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), is_referencevalue(st[r -- i ', refval(r4)][r0 -- i ']), fits(r2, sks1, st[r -- i ', refval(r4)]), is_arraytype(st[r0 -- _type] .type) → i < st[r0 -- _length] .intval ⊦ sks0 = [], r1 = jvmref, r = r4, r = jvmref, sks0 = i ' ' + sks + i ' ' + sks1, r4 = jvmref, r0 = jvmref, hasSharing(r, st), hasSharing(r4, st), junkref(str, r, st), is_classtype(st[r0 -- _type] .type), ∃ 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), ∃ sks3, sks4. sks = sks3 + i ' ' + sks4 ∧ st[r4 -- sks3] .refval = r ∧ (∀ sks2. sks2 + i ' ' ⊑ sks3 → st[r4 -- sks2] .refval ≠ st[r4 -- sks3] .refval)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 272:
(st[r -- i ', refval(r4)][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)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : javavalue, , test-19, ⊦ is_referencevalue(refval(r))])]]