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Proof step 6 from proof k-path (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: path-one: ⊦ path(r ', H) ↔ r ∈ H ∧ r ≠ null;% ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ x0 + [] = x0; (s)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]