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Proof step 689 from proof hasSharing-put-index-refval (previous step)

st[r -- sks3] = refval(r0) ∧ is_referencevalue(st[r -- sks3]), st[r -- sks] = refval(r1), st[r0 -- _type] .type = mkarraytype(ty), (st[r0 -- i ', refval(r2)][(st[r0 -- i ', refval(r2)][r -- (sks3 + i ' ' + sks0)]) .refval -- (i ' ' + sks2)]) = refval(r1), st[r -- (sks3 + i ' ' + sks0)] = refval(r0), fits(r, sks, st), disjoint(str, r ' + r2 ', st), validref(str, r2, ty, st), i < st[r0 -- _length] .intval, 0 ≤ i, fits(r2, sks2, st[r0 -- i ', refval(r2)]), fits(r, sks3 + i ' ' + sks0, st), 0 < st[r0 -- _length] .intval, r2 = jvmref → is_reftype(ty), ∀ sks1. sks1 + i ' ' ⊑ sks3 → st[r -- sks1] .refval ≠ r0 ⊦ r0 = jvmref, r = r2, r1 = jvmref, sks = [], sks3 + (i ' ' + sks0) + i ' ' + sks2 = sks, junkref(str, r, st), hasSharing(r2, st), hasSharing(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), ∃ sks1, sks5. sks3 + i ' ' + sks0 = sks1 + i ' ' + sks5 ∧ st[r -- sks1] .refval = r0 ∧ (∀ sks4. sks4 + i ' ' ⊑ sks1 → st[r -- sks4] .refval ≠ st[r -- sks1] .refval), ∃ sks1. ¬ (sks1 + i ' ' ⊑ sks3 + i ' ' + sks0 → st[r -- sks1] .refval ≠ r0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 688:
st[r -- (sks3 + i ' ' + sks0)] = refval(r0), (st[r0 -- i ', refval(r2)][(st[r0 -- i ', refval(r2)][r -- (sks3 + i ' ' + sks0)]) .refval -- (i ' ' + sks2)]) = refval(r1), st[r0 -- _type] .type = mkarraytype(ty), st[r -- sks] = refval(r1), st[r -- sks3] = refval(r0), 0 < st[r0 -- _length] .intval, fits(r, sks3 + i ' ' + sks0, st), fits(r2, sks2, st[r0 -- i ', refval(r2)]), 0 ≤ i, i < st[r0 -- _length] .intval, validref(str, r2, ty, st), disjoint(str, r ' + r2 ', st), fits(r, sks, st), ∀ sks1. sks1 + i ' ' ⊑ sks3 → st[r -- sks1] .refval ≠ r0, r2 = jvmref → is_reftype(ty) ⊦ sks3 + (i ' ' + sks0) + i ' ' + sks2 = sks, sks = [], r1 = jvmref, r = r2, r0 = jvmref, hasSharing(r, st), hasSharing(r2, st), junkref(str, r, st), ∃ sks1. ¬ (sks1 + i ' ' ⊑ sks3 + i ' ' + sks0 → st[r -- sks1] .refval ≠ r0), ∃ sks1, sks5. sks3 + i ' ' + sks0 = sks1 + i ' ' + sks5 ∧ st[r -- sks1] .refval = r0 ∧ (∀ sks4. sks4 + i ' ' ⊑ sks1 → st[r -- sks4] .refval ≠ st[r -- sks1] .refval), ∃ sks0, sks1. sks = sks0 + i ' ' + sks1 ∧ st[r -- sks0] .refval = r0 ∧ (∀ sks2. sks2 + i ' ' ⊑ sks0 → st[r -- sks2] .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])]]