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Proof step 2 from proof sw-02 (previous step)

∀ a, b. (x0 -1l a) -1l b = (x0 -1l b) -1l a ⊦ ((a0 + x0) -1l a) -1l b = ((a0 + x0) -1l b) -1l a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a, b. (x0 -1l a) -1l b = (x0 -1l b) -1l a ⊦ ((a0 ' + x0) -1l a) -1l b = ((a0 ' + x0) -1l b) -1l a

Rule argument

Additional information

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