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Proof step 2 from proof i2n-divtwo (previous step)

i→n(2 * i) = m0 * 2 + n0 ∧ n0 < 2 ∧ 2 ≠ 0 ⊦ m0 = i→n(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i→n(2 * i) = m0 * 2 + n0, n0 < 2 ⊦ m0 = i→n(i)

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1