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Proof step 8 from proof prefixn-prefixn (previous step)

# y0 = (# y < m ⊃ # y;m), (m ≤ # x ⊃ firstn(m, x);x) = y + z, x = y0 + z0, (# y < m ⊃ # y;m) ≤ # x ⊦ y = y0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# y0 = (# y < m ⊃ # y;m), (m ≤ # y0 + # z0 ⊃ firstn(m, y0 + z0);y0 + z0) = y + z ⊦ y = y0

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)