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Proof step 205 from proof rem-stays-reachable-aux2 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 206:
(x2.last ' + x0).last = ((x2.last ' + x0).last ' + x).last, x + H[x2.last].nxt ' = x2 + r1 ' + x3, (x2.last ' + x0).last ∈ H, path(x2.last ' + x0, H), x2.last ∈ H, H[x2.last].nxt ∈ H, path((x2.last ' + x0).last ' + x2, H), path((x2.last ' + x0).last ' + x2 + r1 ' + x3, 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), (x2.last ' + x0).last ∈ x2, dups(x0), x2.last ∈ x0, x2.last ∈ x

Rule argument

Additional information

Used simplifier rules: path-one: ⊦ path(r ', H) ↔ ¬ ¬(r ∈ H ∧ r ≠ null);%e-01: ⊦ ¬ dups([]); from specification list-dup[reflist]%in-empty: ⊦ ¬ r ∈ []; from specification list[reflist]%l: ⊦ r '.last = r; from specification list-last[reflist]%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]%app: ⊦ x0 + [] = x0; from specification list[reflist]