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Proof step 7 from proof postfix-rec-right (previous step)

∃ z. z + x = a ' + y ⊦ x = a ' + y, ∃ z. z + x = y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
z + x = a ' + y, a ∈ z ⊦ z = [], ∃ z. z + x = y

Rule argument

Additional information

Used simplifier rules: in-append0-not-left: ⊦ a ' + x = x0 + x1 → (¬ a ∈ x0 ↔ x0 = []); from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%app-05: ⊦ x = y + x ↔ 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