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Proof step 21 from proof nextnli-lemma (previous step)

m0 +1 ≤ n1, n1 ≤ n0, m0 +1 ≤ n0, ∀ n. lfgen(n)(0) = x ∧ (∀ m. m < n → lfgen(n)(m + 1) = nextnli(lfgen(n)(m), m)), ∀ n0, n1, lfgen, x, nextnli. m0 ≤ n0 ∧ n1 ≤ n0 ∧ m0 ≤ n1 ∧ (∀ n. lfgen(n)(0) = x ∧ (∀ m. m < n → lfgen(n)(m + 1) = nextnli(lfgen(n)(m), m))) → lfgen(n1)(m0) = lfgen(n0)(m0) ⊦ lfgen(n1)(m0 +1) = lfgen(n0)(m0 +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
m0 < n1, m0 < n0, n1 ≤ n0, ∀ n. lfgen(n)(0) = x ∧ (∀ m. m < n → lfgen(n)(m + 1) = nextnli(lfgen(n)(m), m)), ∀ n0, n1, lfgen, x, nextnli. m0 ≤ n0 ∧ n1 ≤ n0 ∧ m0 ≤ n1 ∧ (∀ n. lfgen(n)(0) = x ∧ (∀ m. m < n → lfgen(n)(m + 1) = nextnli(lfgen(n)(m), m))) → lfgen(n1)(m0) = lfgen(n0)(m0) ⊦ lfgen(n1)(m0 + 1) = lfgen(n0)(m0 + 1)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%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 ≤ n ↔ m < n; (s)% ⊦ m +1 = m + 1; (s)