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Proof step 5 from proof k-path (previous step)

x1 = r0 + x ⊦ path(r ' + x1, H) ↔ 1 + # x1 = 1 ∧ r ∈ H ∧ r ≠ null ∨ 1 = # x1 ∧ r ∈ H ∧ r0 ∈ H ∧ r ≠ null ∧ r0 ≠ null ∧ H[r].nxt = r0 ∨ 2 ≤ 1 + # x1 ∧ r ∈ H ∧ r ≠ null ∧ H[r].nxt = r0 ∧ path(x1, H), x1 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ 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)

Rule argument

Additional information

Used simplifier rules: path-cons: ⊦ path(r ' + r0 ' + x, H) ↔ r ∈ H ∧ r ≠ null ∧ H[r].nxt = r0 ∧ path(r0 ' + x, H);% ⊦ # x0 = 0 ↔ x0 = []; (s)% ⊦ 1 = m + 1 ↔ m = 0; (s)%n < n0 ⊦ (* n) ≠ m0 + (* n0); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ # r ' = 0 +1; (s)% ⊦ #(x0 + x1) = # x0 + # x1; (s)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]% ⊦ r ' + x0 ≠ []; (s)% ⊦ r + x0 = r ' + x0; (s)