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Proof step 3 from proof l-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# x + # y0 = #(y0 -1l a), ∀ x, a. # x + # y0 = #(y0 -1l a) → x = [] ∧ ¬ a ∈ y0 ⊦ a = a0, ¬ a ∈ y0 ∧ x = []

Rule argument

Additional information

Used simplifier rules: del1-y: ⊦ (a ' + x) -1l a = x;%del1-n: ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a);% ⊦ m + n = m + n0 ↔ n = n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)% ⊦ m = m + n ↔ n = 0; (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)