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Proof step 8 from proof isprefix-eq-size-app-eq (previous step)

# x0 = # x1 + # x2, x0 + z = x1 + x2 ⊦ # x0 = #(x1 + x2) ∧ (∀ n. n < # x0 → x0[n] = (x1 + x2)[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# x0 = # x1 + # x2, x0 + z = x1 + x2, n < # x0 ⊦ x0[n] = (x1 + x2)[n], x0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(x + y) = # x + # y; (s)