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Proof step 3 from proof plus (previous step)

2 * m = 2 * m0 + n ⊦ 2 * (m - m0) = n

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 4:
⊦ 2 * m = 2 * n + m0 → ¬ m < n
Proof step 5:
2 * m = 2 * m0 + n → ¬ m < m0, 2 * m = 2 * m0 + n ⊦ 2 * (m - m0) = n

Rule argument

spec: nat-mult inst: name: two-succ-06 seq: ⊦ 2 * m = 2 * n + m0 → ¬ m < n subst: [m, n, m0] → [m, m0, n] flags:

Additional information

Used terms: m, m0, n