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Proof step 40 from proof rem-stays-reachable-aux2 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 41:
⊦ x0 ≠ [] → (x1 + x0).last = x0.last
Proof step 42:
(r ' + x1).last = (r ' + x).last, (r0 ' + x0).last = (r ' + x).last, r0 = H[r0].nxt, x = [], path(r ' + x1, H), path(r0 ' + x0, H), path(r ' + x + r0 ', H), r0 ∈ H, (r ' + x).last ∈ H ⊦ r0 = r, r0 = null, dups(x0), dups(x1), dups(x), r0 ∈ x, r0 ∈ x0, r ∈ x1, r ∈ x
Proof step 52:
(r ' + x1).last = x.last, (r0 ' + x0).last = x.last, r0 = H[r0].nxt, path(r ' + x1, H), path(r0 ' + x0, H), path(r ' + x + r0 ', H), r0 ∈ H, x.last ∈ H ⊦ r0 = r, r0 = null, x = [], dups(x0), dups(x1), dups(x), r0 ∈ x, r0 ∈ x0, r ∈ x1, r ∈ x

Rule argument

spec: list-last, inst: reflist name: lapp seq: ⊦ x0 ≠ [] → (x1 + x0).last = x0.last subst: [x0, x1] → [x, r '] flags: <all paths>

Additional information

Used terms: x, r '