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Proof step 2 from proof one-01 (previous step)

m + n ≤ 1 ⊦ m = 0, m = 1 ∧ n = 0

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: m

Additional information

Used simplifier rules: succ-01: ⊦ m + 1 ≤ n ↔ m < n;%nat freely generated by 0, +1;;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%one-03: ⊦ 1 = m + 1 ↔ m = 0; from specification nat-basic2