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Proof step 2 from proof prefixn-eq-append (previous step)

⊦ prefixn(n, x) = prefixn(n, x + y) ↔ n ≤ # x, y = []

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:
n ≤ # x ⊦ prefixn(n, x + y) = prefixn(n, x)
Proof step 5:
x = (n ≤ # x + # y ⊃ firstn(n, x + y);x + y), # x < n ⊦ y = []

Rule argument

Additional information

Used simplifier rules: prefixn: ⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x);%prefixn-append-left: n ≤ # x ⊦ prefixn(n, x + y) = prefixn(n, x);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(x + y) = # x + # y; (s)