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Proof step 93 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 .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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 92:
st1 .refs ∩ st .refs = ∅, st1[r -- i '] = refval(r0), r ∈ st, st[r -- _type] .type ≤ ty, is_array(str, r, st1), i < st1[r -- _length] .intval, junkrefs(str, r0 ' + acc, st[r -- _type] .type .type, st1 ∪ st), r ∈ st1 .refs, st1[r -- _type] .type ≤ ty, 0 ≤ i, r -- _type ∈ st, 0 < st1[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 = ∅, junkrefs(str, r0 ' + acc, st1[r -- _type] .type .type, st1), r -- _length ∈ st, r -- i ' ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ st .refs = ∅ ↔ st = ∅], [CSIMPFORWARD : ⊦ r ∈ rset → rset ≠ ∅], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ rset ∩ rset0 = rset0 ∩ rset], [CSIMPSEQ : ⊦ (rset ∩ rset0) ∩ rset1 = rset ∩ rset0 ∩ rset1])]]