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Proof step 3 from proof from-from (previous step)

⊦ ⌜x⌝ from m from n = ⌜x⌝ from m + n, # ⌜x⌝ < ⌜m⌝ + ⌜n⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n + m ≤ # x ⊦ x from m from n = x from m + n

Rule argument

Additional information

Used simplifier rules: from-fin: ⊦ ⌜x⌝ from n = ⌜x from n⌝ ;%size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ;%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%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%ls-natinf-to-nat-1-2: ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; from specification nat-inf%inj-02: ⊦ ⌜x⌝ = ⌜x0⌝ ↔ x = x0; from specification stream-basic