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Proof step 6 from proof sw-02 (previous step)

a0 = a, ∀ 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 7:
∀ a, b. (x0 -1l a) -1l b = (x0 -1l b) -1l a ⊦ x0 -1l b = ((a ' + x0) -1l b) -1l a

Rule argument

Additional information

Used simplifier rules: del1-y: ⊦ (a ' + x) -1l a = x;%a: ⊦ (x + y) + z = x + y + z; from specification list