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Proof step 3 from proof emtpystream-merge-to (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n0 < m + n ⊦ ((σ⊥ to n) + (σ⊥ to m))[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)% ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; (s)%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⌝ + ⌜n1⌝ ↔ n + m = n0 + n1; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)%append-assoc: ⊦ (σ0 + σ3) + σ4 = σ0 + σ3 + σ4; from specification stream[buffer-inf]%¬ # σ0 < ⌜n⌝ ⊦ #(σ0 to n) = ⌜n⌝ ; (s)% ⊦ ¬ ∞ < n∞; (s)% ⊦ #(σ0 + σ2) = # σ0 + # σ2; (s)