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Proof step 4 from proof k-path (previous step)

⊦ path(r ' + x1, H) ↔ 1 + # x1 = 1 ∧ r ∈ H ∧ r ≠ null ∨ 1 = # x1 ∧ r ∈ H ∧ hd x1 ∈ H ∧ r ≠ null ∧ hd x1 ≠ null ∧ H[r].nxt = hd x1 ∨ 2 ≤ 1 + # x1 ∧ r ∈ H ∧ r ≠ null ∧ H[r].nxt = hd x1 ∧ path(x1, H)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
x1 = r0 + x ⊦ path(r ' + x1, H) ↔ 1 + # x1 = 1 ∧ r ∈ H ∧ r ≠ null ∨ 1 = # x1 ∧ r ∈ H ∧ r0 ∈ H ∧ r ≠ null ∧ r0 ≠ null ∧ H[r].nxt = r0 ∨ 2 ≤ 1 + # x1 ∧ r ∈ H ∧ r ≠ null ∧ H[r].nxt = r0 ∧ path(x1, H), x1 = []

Rule argument

spec: list-data, inst: reflist name: elim-list-c seq: x0 ≠ [] ⊦ r = hd x0 ∧ x1 = tl x0 ↔ x0 = r + x1 subst: [x0, r, x1] → [x1, r0, x]

Additional information

Used simplifier rules: path-one: ⊦ path(r ', H) ↔ r ∈ H ∧ r ≠ null;% ⊦ x0 + [] = x0; (s)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]% ⊦ # [] = 0; (s)%x0 ≠ [] ⊦ r = hd x0 ∧ x1 = tl x0 ↔ x0 = r + x1; (elim) Used terms: x1, r0, x