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Proof step 11 from proof fill-stream-append (previous step)

m + n = # buf, n0 < m + n ⊦ fill(buf, zero, n, m)[n0] = ((⌜buf⌝ to n) + (σ⊥ to m))[n0], ⌜n0⌝ < #(⌜buf⌝ to n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
m + n = # buf, n0 < m + n, n ≤ n0 ⊦ fill(buf, zero, n, m)[n0] = zero, m = 0

Rule argument

Additional information

Used simplifier rules: zerostream-size: ⊦ # σ⊥ = ∞;% ⊦ m < m + n ↔ n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ (σ⊥)[n] = zero; (s)%m < n, ¬ # σ0 < ⌜n⌝ ⊦ (σ0 to n)[m] = σ0[m]; (s)% ⊦ ¬ ∞ < n∞; (s)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%append-assoc: ⊦ (σ0 + σ3) + σ4 = σ0 + σ3 + σ4; from specification stream[buffer-inf]% ⊦ ⌜m⌝ .n = m; (s)%¬ # σ0 < ⌜n⌝ ⊦ #(σ0 to n) = ⌜n⌝ ; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)% ⊦ m < n → n ≠ 0; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ # ⌜buf⌝ = ⌜# buf⌝ ; (s)% ⊦ ¬ ⌜m⌝ < # σ0 → (σ0 + σ2)[m] = σ2[m - (# σ0).n]; (s)