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Proof step 10 from proof validref-refs-last-valid-subset (previous step)

st0[r0 -- i '] = refval(r2), acc + r1 ' = r2 ' + refs, 0 < st0[r0 -- _length] .intval, 0 ≤ i, validrefs(str, r2 ' + refs, st0[r0 -- _type] .type .type, st0), i < st0[r0 -- _length] .intval, is_array(str, r0, st0), st ⊆ st0, validref(str, r0, st[r0 -- _type] .type, st), st0[r0 -- _type] .type ≤ ty, ∀ str, r, st, st0, r1, ty. validref(str, r, st[r -- _type] .type, st) ∧ st ⊆ st0 ∧ r ∈ acc ∧ validrefs(str, acc + r1 ', ty, st0) → validref(str, r1, st[r1 -- _type] .type, st) ⊦ r0 -- _type ∈ st, r0 = jvmref, validref(str, r1, st[r1 -- _type] .type, st), ¬ r2 ∈ acc ∧ r2 ≠ r1, r1 ≠ r2 ∧ ¬ r1 ∈ refs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
st0[r0 -- i '] = refval(r2), acc + r1 ' = r2 ' + refs, st0[r0 -- _type] .type ≤ ty, 0 < st0[r0 -- _length] .intval, 0 ≤ i, validrefs(str, r2 ' + refs, st0[r0 -- _type] .type .type, st0), i < st0[r0 -- _length] .intval, is_array(str, r0, st0), st ⊆ st0, validref(str, r0, noval .type, st), ∀ str, r, st, st0, r1, ty. validref(str, r, st[r -- _type] .type, st) ∧ st ⊆ st0 ∧ r ∈ acc ∧ validrefs(str, acc + r1 ', ty, st0) → validref(str, r1, st[r1 -- _type] .type, st) ⊦ r0 = jvmref, r0 -- _type ∈ st, validref(str, r1, st[r1 -- _type] .type, st), ¬ r2 ∈ acc ∧ r2 ≠ r1, r1 ≠ r2 ∧ ¬ r1 ∈ refs

Rule argument

[emptyarg]

Additional information

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