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Proof step 1 from proof nextl-lemma-butlast (conclusion)

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

User applied rule cut formula

And obtained the following new goals

Proof step 2:
∀ n. ∃ lf. lf(0) = a ' ∧ (∀ m. m < n → lf(m + 1) = nextl(m, lf(m).last)) ⊦ ∃ lf. lf(0) = a ' ∧ (∀ n. lf(n + 1) = nextl(n, lf(n).last))
Proof step 32:
⊦ ∀ n. ∃ lf. lf(0) = a ' ∧ (∀ m. m < n → lf(m + 1) = nextl(m, lf(m).last)), ∃ lf. lf(0) = a ' ∧ (∀ n. lf(n + 1) = nextl(n, lf(n).last))

Rule argument

fma: ∀ n. ∃ lf. lf(0) = a ' ∧ (∀ m. m < n → lf(m + 1) = nextl(m, lf(m).last))

Additional information

Used formulas: ∀ n. ∃ lf. lf(0) = a ' ∧ (∀ m. m < n → lf(m + 1) = nextl(m, lf(m).last))