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Proof step 10 from proof nextl-lemma-butlast (previous step)

lfgen(0)(0) = a ', ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → lfgen(n1)(m) = lfgen(n2)(m), ∀ n. lfgen(n)(0) = a ' ∧ (∀ m. m < n → lfgen(n)(m + 1) = nextl(m, lfgen(n)(m).last)) ⊦ lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last)

User applied rule all left

And obtained the following new goals

Proof step 11:
lfgen(0)(0) = a ', n ≤ n ∧ n ≤ n + 1 → lfgen(n)(n) = lfgen(n + 1)(n), ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → lfgen(n1)(m) = lfgen(n2)(m), ∀ n. lfgen(n)(0) = a ' ∧ (∀ m. m < n → lfgen(n)(m + 1) = nextl(m, lfgen(n)(m).last)) ⊦ lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last)

Rule argument

pos: left 2 terms: n, n, n + 1 abort: F discard: F computed: F

Additional information

Used terms: n, n, n + 1