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Proof step 96 from proof rem-stays-reachable-aux2 (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 97:
x = x2 + r1 ', (r2 ' + x0).last = r1, x1 + r2 ' = x + H[r2].nxt ' + r0 ' + y, (r ' + x1).last = r1, path(r2 ' + x0, H), r2 ∈ H, H[r2].nxt ∈ H, path(r ' + x1 + r2 ', H), path(r ' + x + H[r2].nxt ', H), path(r ' + x1, H), r1 ∈ H, r2 = H[r2].nxt ∨ r2 = r0 ∨ r2 ∈ y ⊦ H[r2].nxt = null, r = r2, x = [], r2 ∈ x1, r ∈ x1, dups(x0), r2 ∈ x0, r2 ∈ x, dups(x1)

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] → [x, r1, x2] flags: []

Additional information

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