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Proof step 21 from proof odd-add (previous step)

⊦ ∃ m. 2 * m = 2 * m2 + 2 + 2 * m1

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: m2 + 1 + m1 abort: F discard: T computed: F

Additional information

Used simplifier rules: rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%coef-04: ⊦ m * n + m0 * n = m1 * n ↔ ¬ ¬(m + m0 = m1 ∨ n = 0); from specification nat-mult Used terms: m2 + 1 + m1