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Proof step 6 from proof validref-refs-last-valid-subset (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
acc + r1 ' = r2 ' + refs, st0[r0 -- i '] = refval(r2), st0[r0 -- _type] .type ≤ ty, validref(str, r, st[r -- _type] .type, st), st ⊆ st0, is_array(str, r0, st0), i < st0[r0 -- _length] .intval, validrefs(str, r2 ' + refs, st0[r0 -- _type] .type .type, st0), 0 ≤ i, 0 < st0[r0 -- _length] .intval, r ∈ acc, ∀ 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, r = r0, validref(str, r1, st[r1 -- _type] .type, st), r1 ≠ r2 ∧ ¬ r1 ∈ refs, ¬ r2 ∈ acc ∧ r2 ≠ r1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]