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Proof step 2 from proof add-n2i-num (previous step)

⊦ 0 ≤ i → n→i(m) + i = n→i(m + i→n(i))

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: n2i-add: ⊦ n→i(m + n) = n→i(m) + n→i(n);%i2i-pos-01: ⊦ 0 ≤ i → n→i(i→n(i)) = i;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2