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Proof step 106 from proof junkrefs-union-disjoint (previous step)

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, ∃ i0. ¬ i0 < 0 ∧ ¬ st1[r -- _length] .intval ≤ i0 ∧ is_referencevalue(st1[r -- i0 ']) ∧ r0 = st1[r -- i0 '] .refval ∧ junkrefs(str, st1[r -- i0 '] .refval ' + acc, st1[r -- _type] .type .type, st1)

User applied rule exists right

And obtained the following new goals

Proof step 105:
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, ¬ i < 0 ∧ ¬ 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

[EXRARG : pos: right 2, [extquanttermlist: (i)abort:F, discard:T, computed:F]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (i)]]