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Proof step 1 from proof square-inject (conclusion)

⊦ ((m + n) * (m + n + 1)) / 2 + n = ((m0 + n0) * (m0 + n0 + 1)) / 2 + n0 ↔ m = m0 ∧ n = n0

User applied rule cut formula

And obtained the following new goals

Proof step 2:
(((m + n) * (m + n + 1)) / 2 + n) * 2 = (((m0 + n0) * (m0 + n0 + 1)) / 2 + n0) * 2 ⊦ ((m + n) * (m + n + 1)) / 2 + n = ((m0 + n0) * (m0 + n0 + 1)) / 2 + n0 ↔ m = m0 ∧ n = n0

Rule argument

fma: (((m + n) * (m + n + 1)) / 2 + n) * 2 = (((m0 + n0) * (m0 + n0 + 1)) / 2 + n0) * 2

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%zero-02: ⊦ m * n = m * n0 ↔ m = 0 ∨ n = n0; from specification nat-mult Used formulas: (((m + n) * (m + n + 1)) / 2 + n) * 2 = (((m0 + n0) * (m0 + n0 + 1)) / 2 + n0) * 2