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Proof step 4 from proof nextl-lemma-butlast-nonempty (previous step)

∀ 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) ≠ [] ⊦ ∃ lf. lf(0) = a ' ∧ (∀ n. lf(n + 1) = nextl(n, lf(n).last) ∧ lf(n) ≠ [])

User applied rule exists right

And obtained the following new goals

Proof step 5:
∀ 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(0)(0) = a ' ∧ (∀ n. lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last) ∧ lfgen(n)(n) ≠ [])

Rule argument

pos: right 1 terms: λ n. lfgen(n)(n) abort: F discard: T computed: F

Additional information

Used terms: λ n. lfgen(n)(n)