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

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

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] flags:

Additional information

Used simplifier rules: path-one: ⊦ path(r ', H) ↔ ¬ ¬(r ∈ H ∧ r ≠ null);%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 ↔ false);%path-one: ⊦ path(r ', H) ↔ ¬ ¬(r ∈ H ∧ r ≠ null);%c: ⊦ r + x0 = r ' + x0; from specification list[reflist]%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]%len-05: ⊦ #(r ' + x0) = (# x0)+1; from specification list[reflist]%succ: ⊦ m +1 = m + 1; from specification nat-basic2%one-03: ⊦ 1 = m + 1 ↔ m = 0; from specification nat-basic2%l: ⊦ # x0 = 0 ↔ x0 = []; from specification list-data[reflist]%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat%e-01: ⊦ r ' + x0 ≠ []; from specification list[reflist]%app: ⊦ x0 + [] = x0; from specification list[reflist]%size-01: ⊦ # [] = 0; from specification list-data[reflist]%x0 ≠ [] ⊦ r = hd x0 ∧ x1 = tl x0 ↔ x0 = r + x1; (elim) Used terms: x1, r0, x