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Proof step 78 from proof rem-stays-reachable-aux2 (previous step)

x2 = (x + H[x2.last].nxt ') + r0 ' + y, (x2.last ' + x0).last = (r ' + x).last, (r1 ' + y0).last = (r ' + x).last, path(r ' + x2 + r1 ' + y0, H), path(r ' + x + H[x2.last].nxt ', H), path(r ' + x2, H), H[x2.last].nxt ∈ H, x2.last ∈ H, path(x2.last ' + x0, H), (r ' + x).last ∈ H ⊦ H[x2.last].nxt = null, x2.last = r, x2 = [], r = r1, r ∈ x2, x2.last ∈ x, x2.last ∈ x0, dups(x0), r ∈ y0, r1 ∈ x2, r1 ∈ y0, dups(x2 + y0), dups(x2), dups(y0)

User applied rule insert elim lemma

And obtained the following new goals

Proof step 79:
x2 = x1 + r2 ', x2 = (x + H[r2].nxt ') + r0 ' + y, (r2 ' + x0).last = (r ' + x).last, (r1 ' + y0).last = (r ' + x).last, path(r ' + x2 + r1 ' + y0, H), path(r ' + x + H[r2].nxt ', H), path(r ' + x2, H), H[r2].nxt ∈ H, r2 ∈ H, path(r2 ' + x0, H), (r ' + x).last ∈ H ⊦ H[r2].nxt = null, r2 = r, x2 = [], r = r1, r ∈ x2, r2 ∈ x, r2 ∈ x0, dups(x0), r ∈ y0, r1 ∈ x2, r1 ∈ y0, dups(x2 + y0), dups(x2), dups(y0)

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, r2, x1] flags: []

Additional information

Used simplifier rules: x0 ≠ [] ⊦ r = x0.last ∧ x1 = x0.butlast ↔ x0 = x1 + r '; (elim) Used terms: x2, r2, x1