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Proof step 16 from proof sublist-prefix (previous step)

x = a + y, m + m0 < # x, (m + 1) + m0 < # x, n0 ≤ m, firstn(n0, restn(m0, y)) ⊑ firstn(m, restn(m0, y)), ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).tail) ⊑ firstn(n1, restn(m0, x).tail) ⊦ x = [], firstn(n0, restn(m0, y).tail) ⊑ firstn(m, restn(m0, y).tail)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
m + m0 < # y, n0 ≤ m, firstn(n0, restn(m0, y)) ⊑ firstn(m, restn(m0, y)), ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).tail) ⊑ firstn(n1, restn(m0, x).tail) ⊦ y = [], firstn(n0, restn(m0, y).tail) ⊑ firstn(m, restn(m0, y).tail)

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + n < m0 → m + n < m0 + n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)