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Proof step 14 from proof i2n-divtwo (previous step)

2 * i = n→i(m0) * 2 + n→i(n0), n0 < 2, 0 ≤ i ⊦ m0 = i→n(i)

User applied rule cut formula

And closed the goal!

Rule argument

fma: n0 = 1

Additional information

Used simplifier rules: n2n: ⊦ i→n(n→i(n)) = n;%twols-01: ⊦ m ≠ 1 → (m < 2 ↔ m = 0); from specification nat-basic1%add-zero: ⊦ i + 0 = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%odd-even: ⊦ 2 * i ≠ 2 * j + 1; from specification int-mult Used formulas: n0 = 1