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Proof step 2 from proof prefixn-shift-right (previous step)

# x0 ≤ n ⊦ prefixn(n, x0 + x1) = 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:
# x0 ≤ n ⊦ (n ≤ # x0 + # x1 ⊃ firstn(n, x0 + x1);x0 + x1) = x0 + (n ≤ # x1 + # x0 ⊃ firstn(n - # x0, x1);x1)

Rule argument

Additional information

Used simplifier rules: prefixn: ⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x);%n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0; (s)% ⊦ 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)