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

¬ n < # x + # x0 ⊦ nf(n - # x + # x0) = (n < # x ⊃ x[n];(n < # x0 + # x ⊃ x0[n - # x];nf(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-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2