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Proof step 2 from proof add-cancel-num-01 (previous step)

⊦ m ≤ n → (⌜n⌝ = ⌜m⌝ + n∞2 ↔ ⌜n - m⌝ = n∞2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⌜n⌝ = ⌜m⌝ + n∞2, m ≤ n ⊦ ⌜n - m⌝ = n∞2

Rule argument

Additional information

Used simplifier rules: eq-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0;%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2;%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add: ⊦ n ≤ m → ((m - n) + n0 = m ↔ n = n0); from specification nat