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Proof step 15 from proof nextl-lemma-butlast-nonempty (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
lfgen(0)(0) = a ', m ≤ n1, n1 ≤ n0, m ≤ n0, ∀ n. lfgen(n)(0) = a ' ∧ (∀ m. m < n → lfgen(n)(m + 1) = nextl(m, lfgen(n)(m).last) ∧ lfgen(n)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [] ⊦ lfgen(n1)(m) = lfgen(n0)(m), lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last) ∧ lfgen(n)(n) ≠ []

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat