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Proof step 19 from proof rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ x → x ≠ y + a ' + z; (s)% ⊦ ¬ a ∈ x → (a ' + x = y + a ' + z ↔ y = [] ∧ z = x); (ws)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%in-13: ⊦ x = y + a ' + z → a ∈ x; (forward) from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list