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Proof step 3 from proof square-inj (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ (m + n0) + n1 = m + n0 + n1
Proof step 5:
⊦ ((m + n) * (m + n + 1)) / 2 + n = ((m0 + n0) * (m0 + n0 + 1)) / 2 + n0 ↔ m = m0 ∧ n = n0

Rule argument

spec: nat-basic2 inst: name: a seq: ⊦ (m + n0) + n1 = m + n0 + n1 subst: [n1, n0, m] → [1, n0, m0] flags: rewrite

Additional information

Used terms: 1, n0, m0