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Proof step 6 from proof firstn-restn-in (previous step)

x = a0 + y, n0 < # x, ∀ x, a. n0 ≤ # x → (a ∈ x ↔ a ∈ firstn(n0, x) ∨ a ∈ restn(n0, x)) ⊦ a ∈ x ↔ a = a0 ∨ a ∈ firstn(n0, y) ∨ a ∈ restn(n0, y), x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n0 ≤ # y, ∀ x, a. n0 ≤ # x → (a ∈ x ↔ a ∈ firstn(n0, x) ∨ a ∈ restn(n0, x)) ⊦ a ∈ y ↔ a ∈ firstn(n0, y) ∨ a ∈ restn(n0, y), a = a0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)