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Proof step 4 from proof from-notin (previous step)

∀ a. frome(x0, a) = [] ∨ a ∈ x0 ⊦ frome(a0 + x0, a) = [], a ∈ a0 + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a. frome(x0, a) = [] ∨ a ∈ x0 ⊦ a = a0, frome(x0, a) = [], a ∈ x0

Rule argument

Additional information

Used simplifier rules: fromE-no: ⊦ a ≠ b → frome(a ' + x, b) = frome(x, b);%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' + x ↔ ¬(a ≠ b ∧ ¬ a ∈ x); (ss)% ⊦ a + x = a ' + x; (s)