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Proof step 6 from proof firstn-prefix (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n ≤ # x0 + 1, ∀ 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%prefix-rec-02: ⊦ b ' + x ⊑ b ' + y ↔ x ⊑ y; from specification list-last%pred-03: ⊦ n -1 = n - 1; from specification nat