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Proof step 10 from proof gcd-multdiv (previous step)

gcd(m, n) * n1 = m ⊦ ∃ n2. gcd(m, n) * n2 = m * n0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: n1 * n0 abort: F discard: T computed: F

Additional information

Used simplifier rules: zero-02: ⊦ m * n = m * n0 ↔ m = 0 ∨ n = n0; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: n1 * n0