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Proof step 6 from proof restn-firstn (previous step)

2 + m0 + n0 + m0 ≤ # x, ∀ n0, x. m0 + n0 + m0 ≤ # x → restn(m0, firstn(n0 + m0, x)) = firstn(n0, restn(m0, x)) ⊦ restn(m0, firstn(n0 + m0, x.rest)) = firstn(n0, restn(m0, x.rest))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: list-data inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [x, a, y] flags:

Additional information

Used simplifier rules: size-01: ⊦ # [] = 0; from specification list-data%le-num-06: n0 < n ⊦ ¬ m + (* n) ≤ (* n0); from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x, a, y