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

m < 4 ⊦ m = 0, m = 1, m = 2, m = 3

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: m

Additional information

Used simplifier rules: 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: ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; from specification nat-basic2%add-num-04: n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; from specification nat%l-num-05: n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1