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Proof step 1 from proof prefixn-neq-empty-plusone (conclusion)

⊦ prefixn(n + 1, x) = [] ↔ x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x)
Proof step 3:
⊦ prefixn(n, []) = []
Proof step 4:
(n < # x ⊃ firstn(n + 1, x);x) = [] ⊦ x = []

Rule argument

Additional information

Used simplifier rules: prefixn: ⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x);%prefixn-empty: ⊦ prefixn(n, []) = [];%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)