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Proof step 71 from proof addarray-disjoint-nosharing-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 70:
vals[i→n(i)] = refval(r2), vals[i→n(i0)] = refval(r2), st[r2 -- sks] = refval(r0), st[r2 -- sks1] = refval(r0), r2 -- sks ∈ st, st ⊆ addarray(r, ty, vals, st), is_referencevalue(addarray(r, ty, vals, st)[r -- (i0 ' ' + sks)]), 0 ≤ i0, fits(r2, sks, st), fits(r2, 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)]), r2 -- sks1 ∈ st ⊦ i = i0, i0 ' ' + sks = i ' ' + sks1, r = jvmref, vals = [], r0 = jvmref, hasSharing(vals .nonnullrefs, st), junkref(str, r2, st), r2 ≠ jvmref ∧ ¬ # vals ≤ i→n(i0) ∧ ¬ # vals ≤ i→n(i)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : int-abs, , absle, ⊦ 0 ≤ i → abs(i) = i], [CSIMPNAMEDSEQ : intnat, , abs-eq, ⊦ i→n(i) = i→n(j) ↔ abs(i) = abs(j)])]]