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

Rule argument

term: y

Additional information

Used simplifier rules: del1-e: ⊦ [] -1l a = [];%list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ # x = 0 ↔ x = []; (s)% ⊦ ¬ a ∈ []; (s)