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Proof step 22 from proof append-assoc (previous step)

⊦ nf(n - # x + # x0) = (n < # x ⊃ x[n];(n < # x0 + # x ⊃ x0[n - # x];nf(n - # x - # x0))), n < # x + # x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
# x + # x0 ≤ n ⊦ nf(n - # x + # x0) = nf(n - # x - # x0)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic