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Proof step 7 from proof in-prefixn-in-shorter-prefix (previous step)

# y = n0, x = y + z, a ∈ x, a ∈ y, n0 ≤ # x, n1 ≤ # x, n0 ≤ n1 ⊦ a ∈ firstn(n1, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a ∈ y, n1 ≤ # y + # z, # y ≤ n1 ⊦ a ∈ firstn(n1, y + z)

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ a ∈ x → a ∈ x + y; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(x + y) = # x + # y; (s)