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Proof step 475 from proof hasSharing-put-index-refval (previous step)

st[r -- sks0] = refval(r0) ∧ is_referencevalue(st[r -- sks0]), (st[r0 -- i ', refval(r4)][r -- (sks2 + i ' ' + sks)]) = refval(r1), st[r -- sks1] = refval(r0), (st[r0 -- i ', refval(r4)][r4 -- sks3]) = refval(r1), st[r0 -- _type] .type = mkarraytype(ty), st[r -- sks2] = refval(r0), sks0 + i ' ' ⊑ sks2, fits(r, sks2 + i ' ' + sks, st), is_referencevalue(st[r0 -- i ', refval(r4)][r0 -- (i ' ' + sks3)]), 0 ≤ i, i < st[r0 -- _length] .intval, fits(r4, sks3, st[r0 -- i ', refval(r4)]), fits(r, sks1, st), disjoint(str, r ' + r4 ', st), validref(str, r4, ty, st), 0 < st[r0 -- _length] .intval, ∀ sks. sks + i ' ' ⊑ sks2 → st[r -- sks] .refval ≠ r0, r4 = jvmref → is_reftype(ty) ⊦ sks2 + i ' ' + sks = sks1 + i ' ' + sks3, sks1 = [], r0 = jvmref, r = r4, r1 = jvmref, junkref(str, r, st), hasSharing(r4, st), hasSharing(r, st), ∃ sks, sks2. sks1 = sks + i ' ' + sks2 ∧ st[r -- sks] .refval = r0 ∧ (∀ sks4. sks4 + i ' ' ⊑ sks → st[r -- sks4] .refval ≠ st[r -- sks] .refval), ∃ sks. ¬ (sks + i ' ' ⊑ sks1 → st[r -- sks] .refval ≠ r0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 474:
st[r -- sks2] = refval(r0), st[r0 -- _type] .type = mkarraytype(ty), (st[r0 -- i ', refval(r4)][r4 -- sks3]) = refval(r1), st[r -- sks1] = refval(r0), (st[r0 -- i ', refval(r4)][r -- (sks2 + i ' ' + sks)]) = refval(r1), st[r -- sks0] = refval(r0), 0 < st[r0 -- _length] .intval, validref(str, r4, ty, st), disjoint(str, r ' + r4 ', st), fits(r, sks1, st), fits(r4, sks3, st[r0 -- i ', refval(r4)]), i < st[r0 -- _length] .intval, 0 ≤ i, is_referencevalue(st[r0 -- i ', refval(r4)][r0 -- (i ' ' + sks3)]), fits(r, sks2 + i ' ' + sks, st), sks0 + i ' ' ⊑ sks2, r4 = jvmref → is_reftype(ty), ∀ sks. sks + i ' ' ⊑ sks2 → st[r -- sks] .refval ≠ r0 ⊦ r1 = jvmref, r = r4, r0 = jvmref, sks1 = [], sks2 + i ' ' + sks = sks1 + i ' ' + sks3, hasSharing(r, st), hasSharing(r4, st), junkref(str, r, st), ∃ sks. ¬ (sks + i ' ' ⊑ sks1 → st[r -- sks] .refval ≠ r0), ∃ sks, sks2. sks1 = sks + i ' ' + sks2 ∧ st[r -- sks] .refval = r0 ∧ (∀ sks4. sks4 + i ' ' ⊑ sks → st[r -- sks4] .refval ≠ st[r -- sks] .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], [CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1])]]