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Proof step 10 from proof rem-stays-reachable-aux (previous step)

ce.nxt = H[H[(r ' + x).last].nxt].nxt, path(r ' + x, H), H[(r ' + x).last].nxt ∈ H[(r ' + x).last, ce], (r ' + x).last ∈ H, H[(r ' + x).last].nxt ∈ H, ∃ x0. path(r ' + x0, H[(r ' + x).last, ce]) ∧ (r ' + x0).last = H[(r ' + x).last].nxt ∧ ¬ dups(r ' + x0) ⊦ H[(r ' + x).last].nxt = null, H[(r ' + x).last].nxt = ce.nxt, dups(x), r ∈ x, ∃ x0. path(r ' + x0 + H[(r ' + x).last].nxt ', H) ∧ (r ' + x).last ≠ r ∧ ¬ (r ' + x).last ∈ x0, ∃ x0. path(H[(r ' + x).last].nxt ' + x0 + H[(r ' + x).last].nxt ', H) ∧ ¬ (r ' + x).last ∈ x0, H[(r ' + x).last].nxt ≠ (r ' + x).last ∧ r = H[(r ' + x).last].nxt

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
(r ' + x0).last = H[(r ' + x).last].nxt, ce.nxt = H[(r ' + x0).last].nxt, path(r ' + x0, H[(r ' + x).last, ce]), path(r ' + x, H), (r ' + x0).last ∈ H, (r ' + x).last ∈ H, (r ' + x0).last ∈ H[(r ' + x).last, ce] ⊦ (r ' + x0).last = ce.nxt, (r ' + x0).last = null, dups(x), dups(x0), r ∈ x, r ∈ x0, ∃ x0. path(H[(r ' + x).last].nxt ' + x0 + H[(r ' + x).last].nxt ', H) ∧ ¬ (r ' + x).last ∈ x0, ∃ x0. path(r ' + x0 + H[(r ' + x).last].nxt ', H) ∧ (r ' + x).last ≠ r ∧ ¬ (r ' + x).last ∈ x0, (r ' + x0).last ≠ (r ' + x).last ∧ r = (r ' + x0).last

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]% ⊦ dups(r ' + x0) ↔ dups(x0) ∨ r ∈ x0; (ws)