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Proof step 5 from proof nodups-butlastn (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ n ≤ # x → butlastn(n, a ' + x) = a ' + butlastn(n, x)
Proof step 7:
dups(butlastn(n, a ' + x0)), n ≤ 1 + # x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦ dups(x0), a ∈ x0, n ≤ # x0
Proof step 13:
dups(a ' + butlastn(n, x0)), n ≤ 1 + # x0, n ≤ # x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦ dups(x0), a ∈ x0

Rule argument

spec: <toplevel> name: blastN-rec-01 seq: ⊦ n ≤ # x → butlastn(n, a ' + x) = a ' + butlastn(n, x) subst: [x, a, n] → [x0, a, n] flags: <all paths>

Additional information

Used terms: x0, a, n