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Proof step 9 from proof prefixn-append-slice (previous step)

# y = n, x = y + z, n ≤ n0, n ≤ # x ⊦ y + (n0 ≤ (# x - n) + n ⊃ firstn(n0 - n, z);z) = (n0 ≤ # x ⊃ firstn(n0, x);x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# y ≤ n0 ⊦ y + (n0 ≤ # z + # y ⊃ firstn(n0 - # y, z);z) = (n0 ≤ # y + # z ⊃ firstn(n0, y + z);y + z)

Rule argument

Additional information

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