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Proof step 6 from proof extension (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

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, a0, y] flags:

Additional information

Used simplifier rules: size-01: ⊦ # [] = 0; from specification list-data%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%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, a0, y