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Proof step 209 from proof rem-stays-reachable-aux2 (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 210:
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

Rule argument

spec: list-last, inst: reflist name: last-elim seq: x0 ≠ [] ⊦ r = x0.last ∧ x1 = x0.butlast ↔ x0 = x1 + r ' subst: [x0, r, x1] → [x1, r0, x4]

Additional information

Used simplifier rules: x0 ≠ [] ⊦ r = x0.last ∧ x1 = x0.butlast ↔ x0 = x1 + r '; (elim) Used terms: x1, r0, x4