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Proof step 25 from proof nextl-nextj-lemma-butlast (previous step)

m0 < n, m0 + 1 < n, ∀ lfgen, a, iigen, n0, nextl, nextj. (∀ 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))) → (∀ n. m0 ≤ n → lfgen(n)(m0) = lfgen(m0)(m0) ∧ iigen(n)(m0) = iigen(m0)(m0)), ∀ 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)) ⊦ n = m0 + 1, n = 0, lfgen(n)(m0 + 1) = lfgen(m0 + 1)(m0 + 1) ∧ iigen(n)(m0 + 1) = iigen(m0 + 1)(m0 + 1)

User applied rule all left

And obtained the following new goals

Proof step 26:
m0 < n, m0 + 1 < n, (∀ 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))) → (∀ n. m0 ≤ n → lfgen(n)(m0) = lfgen(m0)(m0) ∧ iigen(n)(m0) = iigen(m0)(m0)), ∀ 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)) ⊦ n = m0 + 1, n = 0, lfgen(n)(m0 + 1) = lfgen(m0 + 1)(m0 + 1) ∧ iigen(n)(m0 + 1) = iigen(m0 + 1)(m0 + 1)

Rule argument

pos: left 3 terms: lfgen, a, iigen, n0, nextl, nextj abort: F discard: T computed: F

Additional information

Used terms: lfgen, a, iigen, n0, nextl, nextj