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Proof step 19 from proof rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
a ' + x = x0 + a0 ' + y + a0 ' + z, a ∈ x0 ⊦ a ∈ x, ∃ a0, x0, y, z. x = x0 + a0 ' + y + a0 ' + z

Rule argument

Additional information

Used simplifier rules: neq: ⊦ ¬ a ∈ x → x ≠ y + a ' + z; from specification list%inapp-02: ⊦ ¬ a ∈ y → (y + c ' + z = a ' + z0 ↔ y = [] ∧ c = a ∧ z = z0); 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%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list