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Proof step 3 from proof add-cancel-01 (previous step)

∀ n∞1, n∞2. ⌜n0⌝ + n∞1 = ⌜n0⌝ + n∞2 → n∞1 = n∞2, ⌜n0 +1⌝ + n∞1 = ⌜n0 +1⌝ + n∞2 ⊦ n∞1 = n∞2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
(⌜n0⌝ + ⌜1⌝ ) + n∞1 = (⌜n0⌝ + ⌜1⌝ ) + n∞2, ∀ n∞1, n∞2. ⌜n0⌝ + n∞1 = ⌜n0⌝ + n∞2 → n∞1 = n∞2 ⊦ n∞1 = n∞2

Rule argument

Additional information

Used simplifier rules: add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ;%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2