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Proof step 1 from proof prefixn-prepend-one (conclusion)

⊦ prefixn(n + 1, a ' + x) = a ' + prefixn(n, 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:
⊦ (n ≤ # x ⊃ a ' + firstn(n, x);a ' + x) = a ' + (n ≤ # x ⊃ firstn(n, x);x)

Rule argument

Additional information

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