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Proof step 3 from proof oc-adjoin (previous step)

⊦ #oc(a, x ++ b) = #oc(a, x), a = b

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ¬ a ∈ x → x ++ a = a + x

Rule argument

spec: list-dup name: adjoin-notin seq: ⊦ ¬ a ∈ x → x ++ a = a + x subst: [a, x] → [b, x] flags: <all paths>

Additional information

Used simplifier rules: oc-n: ⊦ a ≠ b → #oc(a, b ' + x) = #oc(a, x);% ⊦ a + x = a ' + x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ x → x ++ a = x; (s) Used terms: b, x