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Proof step 3 from proof l (previous step)

∀ m1, n, n0. m1 < m → true → (m1 + n < m1 + n0 ↔ n < n0) ⊦ m + n < m + n0 ↔ n < n0

User applied rule structural induction

And obtained the following new goals

Proof step 4:
∀ n, n0. (∀ m1, n, n0. m1 < m0 → true → (m1 + n < m1 + n0 ↔ n < n0)) → (m0 + n < m0 + n0 ↔ n < n0), ∀ m1, n, n0. m1 < m0 +1 → true → (m1 + n < m1 + n0 ↔ n < n0) ⊦ m0 +1 + n < m0 +1 + n0 ↔ n < n0

Rule argument

indvar: m

Additional information

Used simplifier rules: z: ⊦ 0 + m = m;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%nat freely generated by 0, +1;;