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Proof step 6 from proof extension (previous step)

# x = 1 + # y0, # x.tail = # y0 ∧ (n < # y0 → x.tail[n] = y0[n]) → x.tail = y0, ∀ n. ¬ # y0 < n → x[n] = (a ' + y0)[n] ⊦ x = a ' + y0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
# x = 1 + # y0, x = a0 + y, # y = # y0 ∧ (n < # y0 → y[n] = y0[n]) → y = y0, ∀ n. ¬ # y0 < n → x[n] = (a ' + y0)[n] ⊦ x = a ' + y0, x = []

Rule argument

spec: list-data name: elim-list-c seq: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y subst: [x, a, y] → [x, a0, y]

Additional information

Used simplifier rules: ⊦ # [] = 0; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x, a0, y