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Proof step 106 from proof rem-stays-reachable-aux2 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 107:
x0 ≠ [] ⊦ x1 + r ' = x0 ↔ x1 = x0.butlast ∧ r = x0.last
Proof step 108:
x1 + r2 ' + x2 = (x + H[r1].nxt ' + r0 ' + y).butlast, (r ' + x1).last = (r ' + x).last, (r1 ' + x0).last = (r ' + x).last, r1 = (x + H[r1].nxt ' + r0 ' + y).last, path(r ' + x1, H), path(r ' + x1 + r2 ' + x2 + r1 ', H), path(r ' + x + H[r1].nxt ', H), path(r1 ' + x0, H), (r ' + x).last ∈ H, r1 ∈ H, H[r1].nxt ∈ H, r1 = H[r1].nxt ∨ r1 = r0 ∨ r1 ∈ y ⊦ x + H[r1].nxt ' + r0 ' + y = [], H[r1].nxt = null, r = r2, r1 = r, r2 = r1, dups(x0), dups(x1 + x2), dups(x1), dups(x2), r ∈ x2, r ∈ x1, r1 ∈ x0, r1 ∈ x, r2 ∈ x1, r2 ∈ x2, r1 ∈ x1, r1 ∈ x2

Rule argument

spec: list-last, inst: reflist name: splitlasteq seq: x0 ≠ [] ⊦ x1 + r ' = x0 ↔ x1 = x0.butlast ∧ r = x0.last subst: [x0, r, x1] → [x + H[r1].nxt ' + r0 ' + y, r1, 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: x + H[r1].nxt ' + r0 ' + y, r1, x1 + r2 ' + x2