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Proof step 254 from proof hasSharing-put-index-refval (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 253:
st[r -- sks0] = refval(r1), st[r -- _type] .type = mkarraytype(ty), (st[r -- i ', refval(r4)][r4 -- ((sks3 + i ' ' + sks4) + i ' ' + sks1)]) = refval(r1), st[r4 -- (sks3 + i ' ' + sks4)] = refval(r), (st[r -- i ', refval(r4)][r4 -- (sks3 + i ' ' + sks4)]) = refval(r0), st[r4 -- sks3] = refval(r), 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, sks3 + i ' ' + sks4, st[r -- i ', refval(r4)]), fits(r0, i ' ' + sks1, st[r -- i ', refval(r4)]) ⊦ sks0 = [], r1 = jvmref, r = r4, r = jvmref, sks0 = i ' ' + (sks3 + i ' ' + sks4) + 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 ∧ (∀ sks. sks + i ' ' ⊑ sks2 → st[r -- sks] .refval ≠ r), ∃ sks2. ¬ (sks2 + i ' ' ⊑ sks3 + i ' ' + sks4 → st[r4 -- sks2] .refval ≠ r), ∃ sks2. ¬ (sks2 + i ' ' ⊑ sks3 → st[r4 -- sks2] .refval ≠ r)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : javavalue, , test-19, ⊦ is_referencevalue(refval(r))], [CSIMPNAMEDFORWARD : int-abs, , forward-le-05, ⊦ i < j ∧ 0 ≤ i → 0 < j])]]