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Proof step 1 from proof postfix-rec-right (conclusion)

⊦ x ⊒ a ' + y ↔ x = a ' + y ∨ x ⊒ y

User applied rule case distinction

And obtained the following new goals

Proof step 2:
x ⊒ a ' + y ⊦ x = a ' + y ∨ x ⊒ y
Proof step 10:
x = a ' + y ∨ x ⊒ y ⊦ x ⊒ a ' + y

Rule argument

pos: right 1

Additional information