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Proof step 15 from proof reach-n-path-n (previous step)

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

Heuristic elimination applied rule insert elim lemma

And closed the goal!

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-cons: ⊦ path(r ' + r0 ' + x, H) ↔ r ∈ H ∧ r ≠ null ∧ H[r].nxt = r0 ∧ path(r0 ' + x, H);%path-first-in-H: ⊦ path(r ' + x, H) → (r ∈ H ↔ true);%path-one: ⊦ path(r ', H) ↔ r ∈ H ∧ r ≠ null;% ⊦ r + x0 = r ' + x0; (s)% ⊦ #(x0 + x1) = # x0 + # x1; (s)% ⊦ # r ' = 0 +1; (s)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]% ⊦ 1 = m + 1 ↔ m = 0; (s)% ⊦ # x0 = 0 ↔ x0 = []; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ r ' + x0 ≠ []; (s)% ⊦ x0 + [] = x0; (s)% ⊦ # [] = 0; (s)%x0 ≠ [] ⊦ r = hd x0 ∧ x1 = tl x0 ↔ x0 = r + x1; (elim) Used terms: x1, r0, x