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Proof step 22 from proof nextnli-lemma (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 23:
m0 < n1, n1 ≤ n0, m0 < n0, m0 ≤ n1 ∧ n1 ≤ n0 ∧ m0 ≤ n0 ∧ (∀ n. lfgen(n)(0) = x ∧ (∀ m. m < n → lfgen(n)(m + 1) = nextnli(lfgen(n)(m), m))) → lfgen(n1)(m0) = lfgen(n0)(m0), ∀ n. lfgen(n)(0) = x ∧ (∀ m. m < n → lfgen(n)(m + 1) = nextnli(lfgen(n)(m), m)) ⊦ lfgen(n1)(m0 + 1) = lfgen(n0)(m0 + 1), n1 = 0, n0 = 0

Rule argument

pos: left 4 terms: n1, n0, lfgen, x, nextnli abort: F discard: T computed: F

Additional information

Used terms: n1, n0, lfgen, x, nextnli