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Proof step 5 from proof firstrestex (previous step)

x = a + y, n0 + 1 < # x, ∀ x. n0 < # x → x = firstn(n0, x) + restn(n0, x) ∨ x = [] ⊦ x = (a ' + firstn(n0, y)) + restn(n0, y), x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 < # y, ∀ x. n0 < # x → x = firstn(n0, x) + restn(n0, x) ∨ x = [] ⊦ y = firstn(n0, y) + restn(n0, y), y = []

Rule argument

Additional information

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