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Proof step 4 from proof nodups-butlastn (previous step)

n ≤ # x0 + 1, dups(butlastn(n, a ' + x0)), ∀ n0. ¬ n0 ≤ # x0 ∨ ¬ dups(butlastn(n0, x0)) ⊦ a ∈ x0, dups(x0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ n ≤ # x → butlastn(n, a ' + x) = a ' + butlastn(n, x)
Proof step 6:
¬ n ≤ # x0, n ≤ # x0 + 1, dups(butlastn(n, a ' + x0)), ∀ n0. ¬ n0 ≤ # x0 ∨ ¬ dups(butlastn(n0, x0)) ⊦ a ∈ x0, dups(x0)
Proof step 12:
n ≤ # x0, n ≤ # x0 + 1, dups(a ' + butlastn(n, x0)), ∀ n0. ¬ n0 ≤ # x0 ∨ ¬ dups(butlastn(n0, x0)) ⊦ a ∈ x0, dups(x0)

Rule argument

spec: inst: name: blastN-rec-01 seq: ⊦ n ≤ # x → butlastn(n, a ' + x) = a ' + butlastn(n, x) subst: [x, a, n] → [x0, a, n] flags: rewrite, current

Additional information

Used terms: x0, a, n