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Proof step 3 from proof path-split-last (previous step)

∀ r, r0, H. path(r ' + x0 + r0 ', H) ↔ path(r ' + x0, H) ∧ H[(r ' + x0).last].nxt = r0 ∧ H[(r ' + x0).last].nxt ∈ H ∧ H[(r ' + x0).last].nxt ≠ null ⊦ path(r ' + (r1 + x0) + r0 ', H) ↔ path(r ' + (r1 + x0), H) ∧ H[(r ' + (r1 + x0)).last].nxt = r0 ∧ H[(r ' + (r1 + x0)).last].nxt ∈ H ∧ H[(r ' + (r1 + x0)).last].nxt ≠ null

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
r ∈ H, ∀ r, r0, H. path(r ' + x0 + r0 ', H) ↔ path(r ' + x0, H) ∧ H[(r ' + x0).last].nxt = r0 ∧ H[(r ' + x0).last].nxt ∈ H ∧ H[(r ' + x0).last].nxt ≠ null ⊦ r = null, path(H[r].nxt ' + x0 + r0 ', H) ↔ path(H[r].nxt ' + x0, H) ∧ H[(H[r].nxt ' + x0).last].nxt = r0 ∧ H[(H[r].nxt ' + x0).last].nxt ∈ H ∧ H[(H[r].nxt ' + x0).last].nxt ≠ null

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));%l-05: ⊦ (r ' + r0 ' + x0).last = (r0 ' + x0).last; from specification list-last[reflist]%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]%c: ⊦ r + x0 = r ' + x0; from specification list[reflist]