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Proof step 104 from proof rem-stays-reachable-aux2 (previous step)

x1 + r2 ' + x3 = x + H[x3.last].nxt ' + r0 ' + y, (r ' + x1).last = (r ' + x).last, (x3.last ' + x0).last = (r ' + x).last, path(r ' + x1, H), path(r ' + x1 + r2 ' + x3, H), path(r ' + x + H[x3.last].nxt ', H), path(x3.last ' + x0, H), (r ' + x).last ∈ H, x3.last ∈ H, H[x3.last].nxt ∈ H ⊦ x3.last = r, H[x3.last].nxt = null, r = r2, x3 = [], dups(x0), dups(x1 + x3), dups(x1), dups(x3), r ∈ x3, r ∈ x1, x3.last ∈ x0, x3.last ∈ x, r2 ∈ x1, r2 ∈ x3

User applied rule insert elim lemma

And obtained the following new goals

Proof step 105:
x1 + r2 ' + x3 = x + H[r1].nxt ' + r0 ' + y, (r1 ' + x0).last = (r ' + x).last, (r ' + x1).last = (r ' + x).last, x3 = x2 + r1 ', path(r1 ' + x0, H), path(r ' + x + H[r1].nxt ', H), path(r ' + x1 + r2 ' + x3, H), path(r ' + x1, H), H[r1].nxt ∈ H, r1 ∈ H, (r ' + x).last ∈ H ⊦ H[r1].nxt = null, r = r2, r1 = r, x3 = [], dups(x3), dups(x1), dups(x1 + x3), dups(x0), r2 ∈ x3, r2 ∈ x1, r1 ∈ x, r1 ∈ x0, r ∈ x1, r ∈ x3

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] → [x3, r1, x2]

Additional information

Used simplifier rules: x0 ≠ [] ⊦ r = x0.last ∧ x1 = x0.butlast ↔ x0 = x1 + r '; (elim) Used terms: x3, r1, x2