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Proof step 5 from proof prefixn-prefixn (previous step)

# y = n, (m ≤ # x ⊃ firstn(m, x);x) = y + z, n ≤ # (m ≤ # x ⊃ firstn(m, x);x) ⊦ y = ((n < m ⊃ n;m) ≤ # x ⊃ firstn((n < m ⊃ n;m), x);x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
(m ≤ # x ⊃ firstn(m, x);x) = y + z ⊦ y = ((# y < m ⊃ # y;m) ≤ # x ⊃ firstn((# y < m ⊃ # y;m), x);x)

Rule argument

Additional information

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