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Proof step 19 from proof refine-add (previous step)

xs1 + prefixn(1, ys) = prefixn(# xs1, xs0) + prefixn(1, dropn(# xs1, xs0)), # xs1 + # ys ≤ # xs0, # xs1 ≤ # xs0, ys = [] ∨ # ys = # xs1 + 1 ⊦ xs1 = prefixn(# xs1, xs0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ # x = # z → (x + y = z + y0 ↔ x = z ∧ y = y0)

Rule argument

spec: list name: len-01 seq: ⊦ # x = # z → (x + y = z + y0 ↔ x = z ∧ y = y0) subst: [y0, z, y, x] → [prefixn(1, dropn(# xs1, xs0)), prefixn(# xs1, xs0), prefixn(1, ys), xs1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ # x ≠ # y → x ≠ y; (s)%n ≤ # x ⊦ # prefixn(n, x) = n; (s)%# x ≤ m ⊦ prefixn(m, x) = x; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: prefixn(1, dropn(# xs1, xs0)), prefixn(# xs1, xs0), prefixn(1, ys), xs1