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Proof step 45 from proof nextl-nextj-lemma-butlast (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
⊦ ∃ lf, jj. lf(0) = a ' ∧ jj(0) = n0 ∧ lf(1) = nextl(jj(0), lf(0).last) ∧ jj(1) = nextj(jj(0), lf(0).last)

Rule argument

Additional information

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