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Proof step 210 from proof rem-stays-reachable-aux2 (previous step)

x + H[x2.last].nxt ' = x2 + r1 ' + x3, (x2.last ' + x0).last = r0, r0 = (r ' + x).last, x1 = x4 + r0 ', path(r ' + x2 + r1 ' + x3, H), path(x2.last ' + x0, H), path(r ' + x1, H), path(r ' + x2, H), r0 ∈ 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, x1 = [], 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 211:
x + H[x2.last].nxt ' = x2 + r1 ' + x3, (x2.last ' + x0).last = (r ' + x).last, path(r ' + x2, H), path(r ' + x4 + (r ' + x).last ', H), path(x2.last ' + x0, H), path(r ' + x2 + r1 ' + x3, H), H[x2.last].nxt ∈ H, x2.last ∈ H, (r ' + x).last ∈ H, H[x2.last].nxt ∈ x2 ∨ H[x2.last].nxt = r1 ∨ H[x2.last].nxt ∈ x3 ⊦ x2.last = r, H[x2.last].nxt = null, r = (r ' + x).last, x2 = [], dups(x2), dups(x0), dups(x4), x2.last ∈ x, x2.last ∈ x0, r ∈ x4, r ∈ x2, (r ' + x).last ∈ x4

Rule argument

Additional information

Used simplifier rules: ⊦ dups(x0 + r ') ↔ r ∈ x0 ∨ dups(x0); (ws)% ⊦ r ∈ r0 ' ↔ r = r0; (s)% ⊦ r ∈ x0 + x1 ↔ r ∈ x0 ∨ r ∈ x1; (s)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]% ⊦ x0 + r ' ≠ []; (s)