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Proof step 5 from proof prefix-one-01 (previous step)

x = a0 + x0 ⊦ (∃ z. x + z = a ') ↔ x = [] ∨ x = a '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ (∃ z. (a0 ' + x0) + z = a ') ↔ a0 ' + x0 = a '

Rule argument

Additional information

Used simplifier rules: ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)