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Proof step 745 from proof hasSharing-put-index-refval (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 744:
st[r0 -- _type] .type = mkarraytype(ty), st[r -- sks] = refval(r1), st[r -- sks1] = refval(r0), (st[r0 -- i ', refval(r2)][r2 -- (sks0 + i ' ' + sks3)]) = refval(r1), st[r2 -- sks0] = 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, sks0 + i ' ' + sks3, st[r0 -- i ', refval(r2)]) ⊦ sks1 = [], sks1 + i ' ' + sks0 + i ' ' + sks3 = sks, sks = [], r1 = jvmref, r = r2, r0 = jvmref, r2 = jvmref, hasSharing(r, st), hasSharing(r2, st), junkref(str, r, st), ∃ sks1, sks3. sks = sks1 + i ' ' + sks3 ∧ st[r -- sks1] .refval = r0 ∧ (∀ sks2. sks2 + i ' ' ⊑ sks1 → st[r -- sks2] .refval ≠ st[r -- sks1] .refval), ∃ sks3. ¬ (sks3 + i ' ' ⊑ sks1 → st[r -- sks3] .refval ≠ r0), ∃ sks, sks3. sks1 = sks + i ' ' + sks3 ∧ st[r -- sks] .refval = r0 ∧ (∀ sks4. sks4 + i ' ' ⊑ sks → st[r -- sks4] .refval ≠ st[r -- sks] .refval), ∃ sks4. ¬ (sks4 + i ' ' ⊑ sks0 → st[r2 -- sks4] .refval ≠ r0)

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])]]