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Proof step 94 from proof rem-stays-reachable-aux2 (previous step)

x1 + r2 ' = H[r2].nxt ' + r0 ' + y, (x0.last ' + x1).last = x0.last, H[x0.last].nxt = H[r2].nxt, path(r2 ' + x0, H), path(x0.last ' + x1, H), path(x0.last ' + x1 + r2 ', H), x0.last ∈ H, r2 ∈ H, H[r2].nxt ∈ H, r2 = H[r2].nxt ∨ r2 = r0 ∨ r2 ∈ y, H[r2].nxt ∈ x1 ∨ H[r2].nxt = r2 ⊦ x0.last = null, x0.last = r2, H[r2].nxt = null, x0 = [], dups(x1), dups(x0), x0.last ∈ x1, r2 ∈ x0, r2 ∈ x1

User applied rule insert elim lemma

And obtained the following new goals

Proof step 95:
x1 + r2 ' = H[r2].nxt ' + r0 ' + y, (r ' + x1).last = r, H[r].nxt = H[r2].nxt, x0 = x + r ', path(r ' + x1 + r2 ', H), path(r ' + x1, H), path(r2 ' + x0, H), H[r2].nxt ∈ H, r2 ∈ H, r ∈ H, H[r2].nxt ∈ x1 ∨ H[r2].nxt = r2, r2 = H[r2].nxt ∨ r2 = r0 ∨ r2 ∈ y ⊦ H[r2].nxt = null, r = r2, r = null, x0 = [], dups(x0), dups(x1), r2 ∈ x1, r2 ∈ x0, r ∈ x1

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] → [x0, r, x]

Additional information

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