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Proof step 129 from proof hasSharing-put-index-refval (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 128:
(st[r -- i ', refval(r2)][r2 -- sks3]) = refval(r1), st[r -- _type] .type = mkarraytype(ty), (st[r -- i ', refval(r2)][r2 -- ((sks1 + i ' ' + sks4) + i ' ' + sks0)]) = refval(r1), st[r2 -- (sks1 + i ' ' + sks4)] = refval(r), (st[r -- i ', refval(r2)][r2 -- (sks1 + i ' ' + sks4)]) = refval(r0), st[r2 -- sks1] = refval(r), 0 < st[r -- _length] .intval, is_referencevalue(st[r -- i ', refval(r2)][r -- (i ' ' + (sks1 + i ' ' + sks4) + i ' ' + sks0)]), fits(r2, sks1 + i ' ' + sks4, st[r -- i ', refval(r2)]), fits(r0, i ' ' + sks0, st[r -- i ', refval(r2)]), i < st[r -- _length] .intval, 0 ≤ i, fits(r2, sks3, st[r -- i ', refval(r2)]), disjoint(str, r ' + r2 ', st), validref(str, r2, ty, st) ⊦ r = jvmref, r = r2, r1 = jvmref, r2 = jvmref, sks3 = [], (sks1 + i ' ' + sks4) + i ' ' + sks0 = sks3, junkref(str, r, st), hasSharing(r2, st), hasSharing(r, st), ∃ sks2. ¬ (sks2 + i ' ' ⊑ sks1 + i ' ' + sks4 → st[r2 -- sks2] .refval ≠ r), ∃ sks2. ¬ (sks2 + i ' ' ⊑ sks1 → st[r2 -- 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], [CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1])]]