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Proof step 4 from proof add-cancel-01 (previous step)

(⌜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

User applied rule all left

And obtained the following new goals

Proof step 5:
(⌜n0⌝ + ⌜1⌝ ) + n∞1 = (⌜n0⌝ + ⌜1⌝ ) + n∞2, ⌜n0⌝ + n∞1 + ⌜1⌝ = ⌜n0⌝ + n∞2 + ⌜1⌝ → n∞1 + ⌜1⌝ = n∞2 + ⌜1⌝ ⊦ n∞1 = n∞2

Rule argument

pos: left 2 terms: n∞1 + ⌜1⌝ , n∞2 + ⌜1⌝ abort: F discard: T computed: T

Additional information

Used terms: n∞1 + ⌜1⌝ , n∞2 + ⌜1⌝