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Proof step 4 from proof l-02 (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
a ∈ y0, # x + # y0 = #(y0 -1l a), ∀ x, a. # x + # y0 = #(y0 -1l a) → x = [] ∧ ¬ a ∈ y0 ⊦ a = a0, ¬ a ∈ y0 ∧ x = []

Rule argument

fma: a ∈ y0

Additional information

Used simplifier rules: in-02-01: ⊦ ¬ a ∈ x → x -1l a = x;%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used formulas: a ∈ y0