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Proof step 208 from proof rem-stays-reachable-aux2 (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 209:
x1 = x4 + r0 ', x + H[x2.last].nxt ' = x2 + r1 ' + x3, (x2.last ' + x0).last = r0, r0 = (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), r0 ∈ 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, x1 = [], x2.last ∈ x, x2.last ∈ x0, dups(x0), r ∈ x1, dups(x1), r ∈ x2, dups(x2)

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] flags: []

Additional information

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