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

x = r + x1 ⊦ 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), x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ path(r ' + x1, H) ↔ 1 + # x1 = 1 ∧ r ∈ H ∧ r ≠ null ∨ 1 = # x1 ∧ r ∈ H ∧ hd x1 ∈ H ∧ r ≠ null ∧ hd x1 ≠ null ∧ H[r].nxt = hd x1 ∨ 2 ≤ 1 + # x1 ∧ r ∈ H ∧ r ≠ null ∧ H[r].nxt = hd x1 ∧ path(x1, H)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; (s)% ⊦ # 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)