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Proof step 65 from proof validrefs-subset-equiv (previous step)

¬ (r -- i ' ∈ st ∧ st ⊆ st0), st0[r -- i '] = refval(r0), validrefs(str, r0 ' + refs1, st0[r -- _type] .type .type, st0), i < st0[r -- _length] .intval, st0[r -- _type] .type ≤ ty, st ⊆ st0, st[r -- _type] .type ≤ ty, is_array(str, r, st), 0 < st0[r -- _length] .intval, 0 ≤ i, ∀ st, st1, ty, r1, str, refs0. # refs0 < # (r0 + refs1) → (validrefs(str, r1 ' + refs0, ty, st1) ∧ st ⊆ st1 ∧ validref(str, r1, st[r1 -- _type] .type, st) → validrefs(str, r1 ' + refs0, ty, st)) ⊦ r = jvmref, junkref(str, r, st), ¬ st[r -- _length] .intval ≤ i ∧ is_referencevalue(st[r -- i ']) ∧ r0 = st[r -- i '] .refval ∧ validrefs(str, st[r -- i '] .refval ' + refs1, st[r -- _type] .type .type, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 64:
st0[r -- i '] = refval(r0), validrefs(str, r0 ' + refs1, st0[r -- _type] .type .type, st0), i < st0[r -- _length] .intval, st0[r -- _type] .type ≤ ty, st ⊆ st0, st[r -- _type] .type ≤ ty, is_array(str, r, st), 0 < st0[r -- _length] .intval, 0 ≤ i, ∀ st, st1, ty, r1, str, refs0. # refs0 < # (r0 + refs1) → (validrefs(str, r1 ' + refs0, ty, st1) ∧ st ⊆ st1 ∧ validref(str, r1, st[r1 -- _type] .type, st) → validrefs(str, r1 ' + refs0, ty, st)) ⊦ r = jvmref, junkref(str, r, st), r -- i ' ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ is_referencevalue(noval)], [CSIMPSEQ : ⊦ ∅[rk] = noval], [CSIMPSEQ : ⊦ ¬ rk ∈ st → st[rk] = ∅[rk]])]]