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Proof step 6 from proof postfix-trans (previous step)

∃ z. z + x = x0, x0 ⊒ y ⊦ ∃ z. z + x = y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
z + x ⊒ y ⊦ ∃ z. z + x = y

Rule argument

Additional information