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Proof step 117 from proof rem-stays-reachable-aux2 (previous step)

(r ' + x1).last = (r ' + x1 + (r2 ' + x2).butlast).last, (r0 ' + x0).last = (r ' + x1).last, H[r0].nxt = (r2 ' + x2).last, x2 = [], path(r ' + x1, H), path(r ' + x1 + r2 ' + x2 + r0 ', H), path(r ' + x1 + r2 ' + x2, H), path(r0 ' + x0, H), (r ' + x1).last ∈ H, r0 ∈ H, H[r0].nxt ∈ H, H[r0].nxt ∈ x1 ∨ H[r0].nxt = r2 ∨ H[r0].nxt ∈ x2 ⊦ H[r0].nxt = null, r = r2, r0 = r, r2 = r0, dups(x0), dups(x1 + x2), dups(x1), dups(x2), r0 ∈ (r2 ' + x2).butlast, r2 ∈ x1, r2 ∈ x2, r0 ∈ x1, r0 ∈ x2, r ∈ x2, r ∈ x1, r0 ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 118:
(r0 ' + x0).last = (r ' + x1).last, path(r ' + x1 + H[r0].nxt ' + r0 ', H), path(r ' + x1 + H[r0].nxt ', H), path(r0 ' + x0, H), path(r ' + x1, H), (r ' + x1).last ∈ H, r0 ∈ H, H[r0].nxt ∈ H ⊦ H[r0].nxt = null, H[r0].nxt = r0, r = H[r0].nxt, r0 = r, dups(x1), dups(x0), H[r0].nxt ∈ x1, r0 ∈ x1, r ∈ x1, r0 ∈ x0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ dups([]); (s)% ⊦ ¬ r ∈ []; (s)% ⊦ r '.butlast = []; (s)% ⊦ r '.last = r; (s)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]% ⊦ x0 + [] = x0; (s)