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Proof step 39 from proof addarray-disjoint-nosharing-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 38:
(addarray(r, ty, vals, st)[r1 -- sks1]) = refval(r0), 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), validreflist(str, vals, ty, st), disjoint(str, vals .nonnullrefs, st), i0 < n→i(# vals), 0 ≤ i, fits(r1, sks1, st), fits(r2, sks, st), 0 ≤ i0, is_referencevalue(addarray(r, ty, vals, st)[r -- (i0 ' ' + sks)]), st ⊆ addarray(r, ty, vals, st), r2 -- sks ∈ st ⊦ r0 = jvmref, vals = [], r = jvmref, i0 ' ' + sks = i ' ' + sks1, junkref(str, r2, st), hasSharing(vals .nonnullrefs, st), dups(vals .nonnullrefs), junkref(str, r1, st), r1 -- sks1 ∈ st

Rule argument

[emptyarg]

Additional information

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