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Proof step 206 from proof rem-stays-reachable-aux2 (previous step)

x + H[x2.last].nxt ' = x2 + r1 ' + x3, (x2.last ' + x0).last = (r ' + x).last, (r ' + x1).last = (r ' + x).last, x1 = [], path(r ' + x2 + r1 ' + x3, H), path(x2.last ' + x0, H), path(r ' + x1, H), path(r ' + x2, H), (r ' + x).last ∈ H, x2.last ∈ H, H[x2.last].nxt ∈ H, H[x2.last].nxt ∈ x2 ∨ H[x2.last].nxt = r1 ∨ H[x2.last].nxt ∈ x3 ⊦ x2.last = r, H[x2.last].nxt = null, x2 = [], dups(x2), dups(x1), dups(x0), r ∈ x2, r ∈ x1, x2.last ∈ x0, x2.last ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 207:
x + H[x2.last].nxt ' = x2 + r1 ' + x3, (x2.last ' + x0).last = ((x2.last ' + x0).last ' + x).last, path((x2.last ' + x0).last ' + x2, H), path((x2.last ' + x0).last ' + x2 + r1 ' + x3, H), path(x2.last ' + x0, H), (x2.last ' + x0).last ∈ H, x2.last ∈ H, H[x2.last].nxt ∈ H, H[x2.last].nxt ∈ x2 ∨ H[x2.last].nxt = r1 ∨ H[x2.last].nxt ∈ x3 ⊦ (x2.last ' + x0).last = null, x2.last = (x2.last ' + x0).last, H[x2.last].nxt = null, x2 = [], dups(x2), dups(x0), (x2.last ' + x0).last ∈ x2, x2.last ∈ x0, x2.last ∈ x

Rule argument

Additional information

Used simplifier rules: path-one: ⊦ path(r ', H) ↔ r ∈ H ∧ r ≠ null;% ⊦ ¬ dups([]); (s)% ⊦ ¬ r ∈ []; (s)% ⊦ r '.last = r; (s)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]% ⊦ x0 + [] = x0; (s)