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Proof step 7 from proof postfix-trans (
previous step
)
x0 ⊒ y, ∃ z. z + x = x0 ⊦ ∃ z. z + x = y
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 8
:
z + x ⊒ y ⊦ ∃ z. z + x = y
Rule argument
Additional information