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Proof step 2 from proof elim-sub (previous step)

n1 +1 ≤ m, ∀ m, n0. n1 ≤ m → (n0 = m - n1 ↔ m = n0 + n1) ⊦ n0 = m - n1 +1 ↔ m = n0 + n1 +1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n1 < m, ∀ m, n0. n1 ≤ m → (n0 = m - n1 ↔ m = n0 + n1) ⊦ n0 = m - n1 + 1 ↔ m = n0 + n1 + 1

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ m + 1 ≤ n ↔ m < n;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)