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Proof step 94 from proof junkrefs-union-disjoint (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 90:
⊦ r ∈ st ↔ r ∈ st .refs
Proof step 93:
st1[r -- i '] = refval(r0), st1 .refs ∩ st .refs = ∅, 0 < st1[r -- _length] .intval, r -- _type ∈ st, 0 ≤ i, st1[r -- _type] .type ≤ ty, r ∈ st1 .refs, junkrefs(str, r0 ' + acc, st[r -- _type] .type .type, st1 ∪ st), i < st1[r -- _length] .intval, is_array(str, r, st1), st[r -- _type] .type ≤ ty, 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, r -- i ' ∈ st, r -- _length ∈ st, junkrefs(str, r0 ' + acc, st1[r -- _type] .type .type, st1)

Rule argument

[speclemmaarg: ,r ∈ st ↔ r ∈ st .refs ,[st, r] → [st1, r] ,F ,storefunsx , ,instore-inrefs ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st1, r)]]