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Proof step 2 from proof diff-empty (previous step)

x0 \ [] = x0 ⊦ (a + x0) \ [] = a + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
x0 \ [] = x0 ⊦ (a ' + x0) \ [] = a ' + x0

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s)