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Proof step 5 from proof fill-stream-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m + n = # buf, n0 < m + n ⊦ ⌜fill(buf, zero, n, m)⌝ [n0] = ((⌜buf⌝ to n) + (σ⊥ to m))[n0]

Rule argument

Additional information

Used simplifier rules: zerostream-size: ⊦ # σ⊥ = ∞;%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf% ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s)%append-assoc: ⊦ (σ0 + σ3) + σ4 = σ0 + σ3 + σ4; from specification stream[buffer-inf]% ⊦ ¬ ∞ < n∞; (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)% ⊦ #(σ0 + σ2) = # σ0 + # σ2; (s)% ⊦ # fill(buf, d, m, n) = # buf; (s)% ⊦ # ⌜buf⌝ = ⌜# buf⌝ ; (s)