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Proof step 22 from proof rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
⊦ a = a0, a ∈ x0, 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%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list