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Proof step 22 from proof rec (previous step)

∃ a0, x0, y, z. x = x0 + a0 ' + y + a0 ' + z ⊦ a ∈ x, ∃ a0, x0, y, z. a ' + x = x0 + a0 ' + y + a0 ' + z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
⊦ a ∈ x0, a ∈ a0 ', a ∈ y, a ∈ z, ∃ a1, x1, y0, z0. a ' + x0 + a0 ' + y + a0 ' + z = x1 + a1 ' + y0 + a1 ' + z0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)