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Proof step 3 from proof min-add-same (previous step)

⊦ (n0 < n1 ⊃ n + n0;n + n1) = n + (n0 < n1 ⊃ n0;n1)

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic