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Proof step 18 from proof reach-reachnext (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
(H[r].nxt ' + x).last = H[(H[r].nxt ' + x1).last].nxt, path(H[r].nxt ' + x, H), path(H[r].nxt ' + x1, H), (H[r].nxt ' + x1).last ∈ H, r ∈ H, ∀ r, x0, H. (r ' + x0).last = H[(r ' + x1).last].nxt ∧ path(r ' + x0, H) ∧ path(r ' + x1, H) ∧ (r ' + x1).last ∈ H → (r ' + x0).last ≠ null ∧ (r ' + x0).last ∈ H ⊦ r = null, (H[r].nxt ' + x).last ≠ null ∧ (H[r].nxt ' + x).last ∈ H

Rule argument

Additional information

Used simplifier rules: path-cons: ⊦ path(r ' + r0 ' + x, H) ↔ r ∈ H ∧ r ≠ null ∧ H[r].nxt = r0 ∧ path(r0 ' + x, H);% ⊦ (r ' + r0 ' + x0).last = (r0 ' + x0).last; (s)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]% ⊦ r + x0 = r ' + x0; (s)