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Proof step 23 from proof nextl-lemma-butlast-nonempty (previous step)

lfgen(0)(0) = a ', m0 < n1, n1 ≤ n0, m0 < n0, ∀ lfgen, a, n1, n0, nextl. lfgen(0)(0) = a ' ∧ m0 ≤ n1 ∧ n1 ≤ n0 ∧ m0 ≤ n0 ∧ (∀ n. lfgen(n)(0) = a ' ∧ (∀ m. m < n → lfgen(n)(m + 1) = nextl(m, lfgen(n)(m).last) ∧ lfgen(n)(m) ≠ [])) ∧ (∀ n, a. nextl(n, a) ≠ []) → lfgen(n1)(m0) = lfgen(n0)(m0), ∀ n. lfgen(n)(0) = a ' ∧ (∀ m. m < n → lfgen(n)(m + 1) = nextl(m, lfgen(n)(m).last) ∧ lfgen(n)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [] ⊦ 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 24:
lfgen(0)(0) = a ', m0 < n1, n1 ≤ n0, m0 < n0, lfgen(0)(0) = a ' ∧ m0 ≤ n1 ∧ n1 ≤ n0 ∧ m0 ≤ n0 ∧ (∀ n. lfgen(n)(0) = a ' ∧ (∀ m. m < n → lfgen(n)(m + 1) = nextl(m, lfgen(n)(m).last) ∧ lfgen(n)(m) ≠ [])) ∧ (∀ n, a. nextl(n, a) ≠ []) → lfgen(n1)(m0) = lfgen(n0)(m0), ∀ n. lfgen(n)(0) = a ' ∧ (∀ m. m < n → lfgen(n)(m + 1) = nextl(m, lfgen(n)(m).last) ∧ lfgen(n)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [] ⊦ lfgen(n1)(m0 + 1) = lfgen(n0)(m0 + 1), n1 = 0, n0 = 0

Rule argument

pos: left 5 terms: lfgen, a, n1, n0, nextl abort: F discard: T computed: F

Additional information

Used terms: lfgen, a, n1, n0, nextl