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Proof step 750 from proof hasSharing-put-index-refval (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 749:
st[r2 -- sks3] = refval(r0), st[r0 -- _type] .type = mkarraytype(ty), st[r -- sks] = refval(r1), st[r -- sks1] = refval(r0), (st[r0 -- i ', refval(r2)][r2 -- (sks3 + i ' ' + sks0)]) = refval(r1), st[r2 -- sks2] = refval(r0), 0 < st[r0 -- _length] .intval, 0 ≤ i, i < st[r0 -- _length] .intval, validref(str, r2, ty, st), disjoint(str, r ' + r2 ', st), fits(r, sks, st), fits(r, sks1, st), fits(r2, sks3 + i ' ' + sks0, st), sks2 + i ' ' ⊑ sks3, ∀ sks. sks + i ' ' ⊑ sks3 → st[r2 -- sks] .refval ≠ r0 ⊦ sks = [], r1 = jvmref, r = r2, r0 = jvmref, r2 = jvmref, sks1 = [], sks1 + i ' ' + sks3 + i ' ' + sks0 = sks, hasSharing(r, st), hasSharing(r2, st), junkref(str, r, st), ∃ sks0, sks1. sks = sks0 + i ' ' + sks1 ∧ st[r -- sks0] .refval = r0 ∧ (∀ sks2. sks2 + i ' ' ⊑ sks0 → st[r -- sks2] .refval ≠ st[r -- sks0] .refval), ∃ sks. ¬ (sks + i ' ' ⊑ sks1 → st[r -- sks] .refval ≠ r0), ∃ sks0, sks. sks1 = sks0 + i ' ' + sks ∧ st[r -- sks0] .refval = r0 ∧ (∀ sks4. sks4 + i ' ' ⊑ sks0 → st[r -- sks4] .refval ≠ st[r -- sks0] .refval)

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])]]