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Proof step 81 from proof junkrefs-union-disjoint (previous step)

¬ ¬ r -- i ' ∈ st, st1 .refs ∩ st .refs = ∅, (st1 ∪ st)[r -- i '] = refval(r0), (st1 ∪ st)[r -- _type] .type ≤ ty, is_array(str, r, st1), i < (st1 ∪ st)[r -- _length] .intval, junkrefs(str, r0 ' + acc, (st1 ∪ st)[r -- _type] .type .type, st1 ∪ st), r ∈ st1, st1[r -- _type] .type ≤ ty, 0 ≤ i, 0 < (st1 ∪ st)[r -- _length] .intval, ∀ st1, st, r, str, ty. st1 .refs ∩ st .refs = ∅ ∧ r ∈ st1 ∧ junkrefs(str, r ' + acc, ty, st1 ∪ st) → junkrefs(str, r ' + acc, ty, st1) ⊦ r = jvmref, ¬ st1[r -- _length] .intval ≤ i ∧ is_referencevalue(st1[r -- i ']) ∧ r0 = st1[r -- i '] .refval ∧ junkrefs(str, st1[r -- i '] .refval ' + acc, st1[r -- _type] .type .type, st1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 80:
st1 .refs ∩ st .refs = ∅, st[r -- i '] = refval(r0), r -- i ' ∈ st, (st1 ∪ st)[r -- _type] .type ≤ ty, is_array(str, r, st1), i < (st1 ∪ st)[r -- _length] .intval, junkrefs(str, r0 ' + acc, (st1 ∪ st)[r -- _type] .type .type, st1 ∪ st), r ∈ st1, st1[r -- _type] .type ≤ ty, 0 ≤ i, 0 < (st1 ∪ st)[r -- _length] .intval, r ∈ st, ∀ st1, st, r, str, ty. st1 .refs ∩ st .refs = ∅ ∧ r ∈ st1 ∧ junkrefs(str, r ' + acc, ty, st1 ∪ st) → junkrefs(str, r ' + acc, ty, st1) ⊦ r = jvmref, ¬ st1[r -- _length] .intval ≤ i ∧ is_referencevalue(st1[r -- i ']) ∧ r0 = st1[r -- i '] .refval ∧ junkrefs(str, st1[r -- i '] .refval ' + acc, st1[r -- _type] .type .type, st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ rk ∈ st0 → (st ∪ st0)[rk] = st0[rk]])]]