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Proof step 5 from proof le-addex (previous step)

m = n ∨ m < n ⊦ n = 0, ∃ n0. m + n0 = n

User applied rule case distinction

And obtained the following new goals

Proof step 6:
m < n ⊦ n = 0, ∃ n0. m + n0 = n

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ m = m + n ↔ n = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic