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Proof step 34 from proof addarray-disjoint-nosharing-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 33:
vals[i→n(i)] = refval(r1), vals[i→n(i0)] = refval(r2), (addarray(r, ty, vals, st)[r2 -- sks]) = refval(r0), (addarray(r, ty, vals, st)[r1 -- sks1]) = refval(r0), st ⊆ addarray(r, ty, vals, st), is_referencevalue(addarray(r, ty, vals, st)[r -- (i0 ' ' + sks)]), 0 ≤ i0, fits(r2, sks, st), fits(r1, sks1, st), 0 ≤ i, i0 < n→i(# vals), disjoint(str, vals .nonnullrefs, st), validreflist(str, vals, ty, st), newref(r, st), i < n→i(# vals), is_referencevalue(addarray(r, ty, vals, st)[r -- (i ' ' + sks1)]) ⊦ i0 ' ' + sks = i ' ' + sks1, r = jvmref, vals = [], r0 = jvmref, junkref(str, r1, st), dups(vals .nonnullrefs), hasSharing(vals .nonnullrefs, st), junkref(str, r2, st), r2 -- sks ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : addarray, , addarray-subset, ⊦ newref(r, st) → ( ¬ st ⊆ addarray (r, ty, vals, st) ↔ false)])]]