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Proof step 5 from proof l-02 (previous step)

# x + # y0 = #(y0 -1l a), a ∈ y0, ∀ x, a. # x + # y0 = #(y0 -1l a) → x = [] ∧ ¬ a ∈ y0 ⊦ a = a0, ¬ a ∈ y0 ∧ x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# x + # y0 = # y0 - 1, a ∈ y0, ∀ x, a. # x + # y0 = #(y0 -1l a) → x = [] ∧ ¬ a ∈ y0 ⊦ a = a0

Rule argument

Additional information

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