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Proof step 3 from proof nextl-nextj-lemma-butlast (previous step)

∀ m. ∃ lf, jj. lf(0) = a ' ∧ jj(0) = n0 ∧ (∀ n. n ≤ m → lf(n + 1) = nextl(jj(n), lf(n).last) ∧ jj(n + 1) = nextj(jj(n), lf(n).last)) ⊦ ∃ lf, jj. lf(0) = a ' ∧ jj(0) = n0 ∧ (∀ n. lf(n + 1) = nextl(jj(n), lf(n).last) ∧ jj(n + 1) = nextj(jj(n), lf(n).last))

User applied rule choice left

And obtained the following new goals

Proof step 4:
∀ m. lfgen(m)(0) = a ' ∧ iigen(m)(0) = n0 ∧ (∀ n. n ≤ m → lfgen(m)(n + 1) = nextl(iigen(m)(n), lfgen(m)(n).last) ∧ iigen(m)(n + 1) = nextj(iigen(m)(n), lfgen(m)(n).last)) ⊦ ∃ lf, jj. lf(0) = a ' ∧ jj(0) = n0 ∧ (∀ n. lf(n + 1) = nextl(jj(n), lf(n).last) ∧ jj(n + 1) = nextj(jj(n), lf(n).last))

Rule argument

pos: left 1

Proof step 3 from proof nextl-nextj-lemma-butlast (previous step)

User applied rule choice left

And obtained the following new goals

Rule argument

Additional information