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Proof step 1 from proof l-01 (conclusion)

⊦ # x = #(x -1l a) ↔ ¬ a ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# x = # x - 1, a ∈ x ⊦

Rule argument

Additional information

Used simplifier rules: l: ⊦ a ∈ x → #(x -1l a) = (# x)-1;%in-02-01: ⊦ ¬ a ∈ x → x -1l a = x;% ⊦ n -1 = n - 1; (s)