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Proof step 167 from proof pathex-put-nopathex (previous step)

¬ (is_array(str, r, st[r0 -- sk, val]) ∧ r ≠ (r1 ' + refs1) .last), st[r -- i '] = refval(r1), 0 ≤ i, 0 < st[r -- _length] .intval, is_array(str, r, st), i < st[r -- _length] .intval, validrefs(str, r1 ' + refs1, st[r -- _type] .type .type, st), path∃(str, r1, r0, st) ∨ junkref(str, r1, st) ∨ path∃(str, r1, (r1 ' + refs1) .last, st[r0 -- sk, val]), ∀ val, r0, sk, st0, r2, str, refs. # refs < # (r1 + refs1) → ( validrefs(str, r2 ' + refs, st0[r2 -- _type] .type, st0) → sk = _type ∨ path∃(str, r2, r0, st0) ∨ junkref(str, r2, st0) ∨ path∃(str, r2, (r2 ' + refs) .last, st0[r0 -- sk, val])) ⊦ sk = _type, r = jvmref, path∃(str, r, r0, st), path∃(str, r, (r1 ' + refs1) .last, st[r0 -- sk, val]), r1 ≠ jvmref ∧ (¬ st[r1 -- _type] .type ≤ st[r -- _type] .type .type ∨ junkref(str, r1, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 166:
st[r -- i '] = refval(r1), validrefs(str, r1 ' + refs1, st[r -- _type] .type .type, st), i < st[r -- _length] .intval, is_array(str, r, st), 0 < st[r -- _length] .intval, 0 ≤ i, path∃(str, r1, r0, st) ∨ junkref(str, r1, st) ∨ path∃(str, r1, (r1 ' + refs1) .last, st[r0 -- sk, val]), ∀ val, r0, sk, st0, r2, str, refs. # refs < # (r1 + refs1) → ( validrefs(str, r2 ' + refs, st0[r2 -- _type] .type, st0) → sk = _type ∨ path∃(str, r2, r0, st0) ∨ junkref(str, r2, st0) ∨ path∃(str, r2, (r2 ' + refs) .last, st0[r0 -- sk, val])) ⊦ r = jvmref, sk = _type, path∃(str, r, (r1 ' + refs1) .last, st[r0 -- sk, val]), path∃(str, r, r0, st), is_array(str, r, st[r0 -- sk, val]) ∧ r ≠ (r1 ' + refs1) .last, r1 ≠ jvmref ∧ (¬ st[r1 -- _type] .type ≤ st[r -- _type] .type .type ∨ junkref(str, r1, st))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]