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Proof step 2 from proof l-02 (previous step)

# x + # y = #(y -1l a) ⊦ x = [] ∧ ¬ a ∈ y

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ x, a. # x + # y0 = #(y0 -1l a) → x = [] ∧ ¬ a ∈ y0, # x + #(a0 + y0) = #((a0 + y0) -1l a) ⊦ x = [] ∧ ¬ a ∈ a0 + y0

Rule argument

term: y

Additional information

Used simplifier rules: del1-e: ⊦ [] -1l a = [];%list freely generated by [], +;;%size-01: ⊦ # [] = 0; from specification list-data%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%in-empty: ⊦ ¬ a ∈ []; from specification list