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Proof step 144 from proof rem-stays-reachable-aux2 (previous step)

(r0 ' + x0).last = (r2 ' + x2.butlast).last, (r ' + x1).last = (r2 ' + x2.butlast).last, H[r0].nxt = x2.last, path(r0 ' + x0, H), path(r ' + x1 + r2 ' + x2, H), path(r ' + x1 + r2 ' + x2 + r0 ', H), path(r ' + x1, H), x2.last ∈ H, r0 ∈ H, (r2 ' + x2.butlast).last ∈ H ⊦ x2.last = null, r = r2, r0 = r, r2 = r0, x2 = [], dups(x2), dups(x1), dups(x1 + x2), dups(x0), r0 ∈ x0, r ∈ x1, r ∈ x2, r0 ∈ x2, r0 ∈ x1, r2 ∈ x2, r2 ∈ x1, r0 ∈ x2.butlast

User applied rule insert elim lemma

And obtained the following new goals

Proof step 145:
(r ' + x1).last = (r2 ' + x).last, (r0 ' + x0).last = (r2 ' + x).last, H[r0].nxt = r1, x2 = x + r1 ', path(r ' + x1, H), path(r ' + x1 + r2 ' + x2 + r0 ', H), path(r ' + x1 + r2 ' + x2, H), path(r0 ' + x0, H), (r2 ' + x).last ∈ H, r0 ∈ H, r1 ∈ H ⊦ r = r2, r0 = r, r1 = null, r2 = r0, x2 = [], dups(x0), dups(x1 + x2), dups(x1), dups(x2), r0 ∈ x, r2 ∈ x1, r2 ∈ x2, r0 ∈ x1, r0 ∈ x2, r ∈ x2, r ∈ x1, r0 ∈ x0

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] → [x2, r1, x]

Additional information

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