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Proof step 3 from proof emptytream-dist-to (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n0 < m ⊦ ((σ⊥ to n) + σ⊥)[n0] = zero

Rule argument

Additional information

Used simplifier rules: zerostream-size: ⊦ # σ⊥ = ∞;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ (σ⊥)[n] = zero; (s)%m < n, ¬ # σ0 < ⌜n⌝ ⊦ (σ0 to n)[m] = σ0[m]; (s)% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ ⌜m⌝ = ⌜m0⌝ ↔ m = m0; (s)%append-assoc: ⊦ (σ0 + σ3) + σ4 = σ0 + σ3 + σ4; from specification stream[buffer-inf]%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% ⊦ ⌜m⌝ + ∞ = ∞; (s)%¬ # σ0 < ⌜n⌝ ⊦ #(σ0 to n) = ⌜n⌝ ; (s)% ⊦ ¬ ∞ < n∞; (s)% ⊦ #(σ0 + σ2) = # σ0 + # σ2; (s)