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Proof step 38 from proof nextl-lemma (previous step)

∀ nextl, a. ∃ lf. lf(0) = nextl(0, a) ∧ (∀ m. m < n0 → lf(m + 1) = nextl(m + 1, lf(m).last)) ⊦ ∃ lf. lf(0) = nextl(0, a) ∧ (∀ m. ¬ n0 < m → lf(m + 1) = nextl(m + 1, lf(m).last))

User applied rule all left

And obtained the following new goals

Proof step 39:
∃ lf. lf(0) = nextl(0, a) ∧ (∀ m. m < n0 → lf(m + 1) = nextl(m + 1, lf(m).last)) ⊦ ∃ lf. lf(0) = nextl(0, a) ∧ (∀ m. ¬ n0 < m → lf(m + 1) = nextl(m + 1, lf(m).last))

Rule argument

pos: left 1 terms: nextl, a abort: F discard: T computed: T

Additional information

Used terms: nextl, a