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Proof step 11 from proof prefixn-prefixn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# y0 = # y, firstn(m, y0 + z0) = y + z, # y < m, m ≤ # y + # z0 ⊦ y = y0, z0 = []

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% ⊦ # x = 0 ↔ x = []; (s)% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic