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Proof step 18 from proof cong-equiv (previous step)

n1 + m1 * m = m0 * m, n2 < m ⊦ n1 = 0, ∃ n0. m * n0 = n1

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 2 terms: m0 - m1 abort: F discard: T computed: F

Additional information

Used terms: m0 - m1