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Proof step 2 from proof prefix-prefixn-exact (previous step)

n = # x0, x0 ⊑ x1 ⊦ prefixn(n, x1) = x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x)
Proof step 4:
x0 ⊑ x1 ⊦ firstn(# x0, x1) = x0

Rule argument

Additional information

Used simplifier rules: prefixn: ⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x);% ⊦ # y < # x → ¬ x ⊑ y; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)