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Proof step 26 from proof addarray-disjoint-nosharing-01 (previous step)

(addarray(r, ty, vals, st)[r1 -- sks1]) = refval(r0), (addarray(r, ty, vals, st)[r2 -- sks]) = refval(r0), vals[i→n(i0)] = refval(r2), vals[i→n(i)] = refval(r1), is_referencevalue(addarray(r, ty, vals, st)[r -- (i ' ' + sks1)]), i < n→i(# vals), newref(r, st), refval(r1) ∈ vals → is_referencevalue(refval(r1)) ∧ validref(str, refval(r1) .refval, ty, st), disjoint(str, vals .nonnullrefs, st), i0 < n→i(# vals), 0 ≤ i, fits(r1, sks1, addarray(r, ty, vals, st)), fits(r2, sks, st), 0 ≤ i0, is_referencevalue(addarray(r, ty, vals, st)[r -- (i0 ' ' + sks)]), st ⊆ addarray(r, ty, vals, st), junkref(str, r1, st) ⊦ r0 = jvmref, vals = [], r = jvmref, i0 ' ' + sks = i ' ' + sks1, junkref(str, r2, st), hasSharing(vals .nonnullrefs, st), dups(vals .nonnullrefs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
(addarray(r, ty, vals, st)[r1 -- sks1]) = refval(r0), (addarray(r, ty, vals, st)[r2 -- sks]) = refval(r0), vals[i→n(i0)] = refval(r2), vals[i→n(i)] = refval(r1), is_referencevalue(addarray(r, ty, vals, st)[r -- (i ' ' + sks1)]), i < n→i(# vals), newref(r, st), disjoint(str, vals .nonnullrefs, st), i0 < n→i(# vals), 0 ≤ i, fits(r1, sks1, addarray(r, ty, vals, st)), fits(r2, sks, st), 0 ≤ i0, is_referencevalue(addarray(r, ty, vals, st)[r -- (i0 ' ' + sks)]), st ⊆ addarray(r, ty, vals, st), junkref(str, r1, st) ⊦ r0 = jvmref, vals = [], r = jvmref, i0 ' ' + sks = i ' ' + sks1, junkref(str, r2, st), hasSharing(vals .nonnullrefs, st), dups(vals .nonnullrefs), refval(r1) ∈ vals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : junkrefs, , validref-junkref, ⊦ junkref(str, r, st) → ¬ validref(str, r, ty, st)], [CSIMPNAMEDSEQ : javavalue, , sel-09, ⊦ refval(r) .refval = r], [CSIMPNAMEDSEQ : javavalue, , test-19, ⊦ is_referencevalue(refval(r))])]]