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Proof step 2 from proof sublist-prefix (previous step)

n1 + m0 < # x, n ≤ n1 ⊦ sublist(m0, n + 1, x) ⊑ sublist(m0, n1 + 1, x) ↔ true

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n1 + m0 < # x, n ≤ n1 ⊦ x = [], firstn(n, restn(m0, x).tail) ⊑ firstn(n1, restn(m0, x).tail)

Rule argument

Additional information

Used simplifier rules: sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));%firstN-rec: ⊦ firstn(n + 1, x) = x.head + firstn(n, x.tail);% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y ⊑ x + z ↔ y ⊑ z; (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)