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Proof step 2 from proof k-path (previous step)

⊦ path(x, H) ↔ # x = 1 ∧ hd x ∈ H ∧ hd x ≠ null ∨ # x = 2 ∧ hd x ∈ H ∧ hd tl x ∈ H ∧ hd x ≠ null ∧ hd tl x ≠ null ∧ H[hd x].nxt = hd tl x ∨ 2 ≤ # x ∧ hd x ∈ H ∧ hd x ≠ null ∧ H[hd x].nxt = hd tl x ∧ path(tl x, H)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
x = r + x1 ∧ x ≠ [] ⊦ path(x, H) ↔ # x = 1 ∧ r ∈ H ∧ r ≠ null ∨ # x = 2 ∧ r ∈ H ∧ hd x1 ∈ H ∧ r ≠ null ∧ hd x1 ≠ null ∧ H[r].nxt = hd x1 ∨ 2 ≤ # x ∧ r ∈ H ∧ r ≠ null ∧ H[r].nxt = hd x1 ∧ path(x1, H)

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] → [x, r, x0] flags:

Additional information

Used simplifier rules: path-empty: ⊦ ¬ path([], H);%size-01: ⊦ # [] = 0; from specification list-data[reflist]%x0 ≠ [] ⊦ r = hd x0 ∧ x1 = tl x0 ↔ x0 = r + x1; (elim) Used terms: x, r, x0