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Proof step 20 from proof append-assoc (previous step)

n < # x + # x0 ⊦ (x + x0)[n] = (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 21:
n < # x + # x0 ⊦ (x + x0)[n] = (n < # x ⊃ x[n];x0[n - # x])

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic