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

∀ m. n→i(m + n0) = n→i(m) + n→i(n0) ⊦ n→i(m + n0 +1) = n→i(m) + n→i(n0 +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ m. n→i(m + n0) = n→i(m) + n→i(n0) ⊦ n→i(m + n0) = n→i(m) + n→i(n0)

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ n→i(m + 1) = n→i(m) + 1;%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i + j = i + i0 ↔ j = i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)