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Proof step 11 from proof isprefix-eq-size-app-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# x0 = # x1 + # x2, n < # x0, ∀ n0. n0 < # x0 → x0[n0] = (x1 + x2)[n0] ⊦ x0[n] = (x1 + x2)[n], x0 = []

Rule argument

Additional information

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