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Proof step 3 from proof k-path (previous step)

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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ 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)

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ r ' + x0 ≠ []; from specification list[reflist]%le-num-04: n0 < n ⊦ (* n) ≤ m0 + (* n0) ↔ n - n0 ≤ m0; from specification nat%add-num-04: n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l: ⊦ # x0 = 0 ↔ x0 = []; from specification list-data[reflist]%one-03: ⊦ 1 = m + 1 ↔ m = 0; from specification nat-basic2%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(r ' + x0) = (# x0)+1; from specification list[reflist]%c: ⊦ r + x0 = r ' + x0; from specification list[reflist]