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Proof step 21 from proof nextl-nextj-lemma-butlast (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: 0 abort: F discard: T computed: T

Additional information

Used simplifier rules: lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: 0