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Proof step 112 from proof rem-stays-reachable-aux2 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 113:
x0 ≠ [] ⊦ x0 = x1 + r ' ↔ x1 = x0.butlast ∧ r = x0.last
Proof step 114:
(r ' + x1).last = (r ' + x).last, (r0 ' + x0).last = (r ' + x).last, H[r0].nxt = (x1 + r2 ' + x2).last, x = (x1 + r2 ' + x2).butlast, path(r ' + x1, H), path(r ' + x1 + r2 ' + x2 + r0 ', H), path(r ' + x1 + r2 ' + x2, H), path(r0 ' + x0, H), (r ' + x).last ∈ H, r0 ∈ H, H[r0].nxt ∈ H, H[r0].nxt ∈ x1 ∨ H[r0].nxt = r2 ∨ H[r0].nxt ∈ x2 ⊦ x1 + r2 ' + x2 = [], H[r0].nxt = null, r = r2, r0 = r, r2 = r0, dups(x0), dups(x1 + x2), dups(x1), dups(x2), r ∈ x2, r ∈ x1, r0 ∈ x0, r0 ∈ x, r2 ∈ x1, r2 ∈ x2, r0 ∈ x1, r0 ∈ x2

Rule argument

spec: list-last, inst: reflist name: splitlasteq-01 seq: x0 ≠ [] ⊦ x0 = x1 + r ' ↔ x1 = x0.butlast ∧ r = x0.last subst: [r, x1, x0] → [H[r0].nxt, x, x1 + r2 ' + x2] flags: <all paths>

Additional information

Used simplifier rules: ⊦ x0 + x1 = [] ↔ x0 = [] ∧ x1 = []; (as)% ⊦ r ' + x0 ≠ []; (s)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist] Used terms: H[r0].nxt, x, x1 + r2 ' + x2