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Proof step 11 from proof i2n-divtwo (previous step)

n0 ≠ 1, 2 * (n→i(m0) + i) + n→i(n0) = 0, n0 = 0, i < 0 ⊦ m0 = i→n(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
n→i(m0) + i = 0, i < 0 ⊦ m0 = i→n(i)

Rule argument

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%two-zero: ⊦ 2 * i = 0 ↔ i = 0; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-zero: ⊦ i + 0 = i; from specification int-basic2