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Proof step 2 from proof sublist-lenrec-01 (previous step)

n < # x ⊦ sublist(0, n + 1, x) = sublist(0, n, x) + x[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < # x ⊦ x.head ' + firstn(n, x.tail) = firstn(n, x) + x[n] ', x = []

Rule argument

Additional information

Used simplifier rules: sub-zerorec-01: n < # x ⊦ sublist(0, n + 1, x) = x.head + sublist(0, n, x.tail);%sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));%restN-zero: ⊦ restn(0, x) = x;% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ x + a = x + a '; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ a + x = a ' + x; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)