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Proof step 4 from proof prefix-prefixn-not-shorter (previous step)

# x1 < # x0, # x1 < n, n < # x0, # x1 + 1 < # x0, firstn(n, x0) ⊑ x1 ⊦ x0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ x ⊑ y ↔ (∃ z. x + z = y)

Rule argument

spec: list-last name: prefix seq: ⊦ x ⊑ y ↔ (∃ z. x + z = y) subst: [y, x] → [x1, firstn(n, x0)] flags: <all paths>

Additional information

Used simplifier rules: ⊦ #(x + y) = # x + # y; (s)% ⊦ n < # x → # firstn(n, x) = n; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: x1, firstn(n, x0)