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Proof step 14 from proof reach-n-path-n (previous step)

# x = 2, x = r + x1, path(x, H) ⊦ x = [], r ≠ null ∧ H[r].nxt ≠ null ∧ r ∈ H ∧ H[r].nxt ∈ H ∧ H[r].nxt = x.last

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
1 = # x1, path(r ' + x1, H) ⊦ r = null, H[r].nxt = hd x1 ∧ H[r].nxt ∈ H ∧ H[r].nxt ≠ null

Rule argument

Additional information

Used simplifier rules: path-first-in-H: ⊦ path(r ' + x, H) → (r ∈ H ↔ true);%null-not-start-of-path: ⊦ ¬ path(null ' + x, H);% ⊦ # x0 = 1 → (x1 + x0).last = hd x0; (s)% ⊦ r ' + x0 ≠ []; (s)%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)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]% ⊦ # r ' = 0 +1; (s)% ⊦ #(x0 + x1) = # x0 + # x1; (s)% ⊦ r + x0 = r ' + x0; (s)