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Proof step 2 from proof prefix-prefixn-not-shorter (previous step)

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

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:
# x1 < # x0, # x1 < n, n < # x0, # x1 + 1 < # x0, firstn(n, x0) ⊑ x1 ⊦ x0 = []

Rule argument

Additional information

Used simplifier rules: prefixn: ⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x);% ⊦ m +1 = m + 1; (s)% ⊦ # x = 0 ↔ x = []; (s)% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → m ≤ n; (s)