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Proof step 7 from proof firstn-prefix (previous step)

n ≤ 1 + # x0, ∀ n. n ≤ # x0 → firstn(n, x0) ⊑ x0 ⊦ n = 0, a ' + firstn(n -1, x0) ⊑ a ' + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n ≤ 1 + # x0, ∀ n. n ≤ # x0 → firstn(n, x0) ⊑ x0 ⊦ n = 0, firstn(n - 1, x0) ⊑ x0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y ⊑ x + z ↔ y ⊑ z; (s)% ⊦ n -1 = n - 1; (s)