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Proof step 10 from proof sw-02 (previous step)

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

Heuristic quantifier closing applied rule all left

And obtained the following new goals

Proof step 11:
(x0 -1l a0) -1l b = (x0 -1l b) -1l a0, ∀ a, b. (x0 -1l a) -1l b = (x0 -1l b) -1l a ⊦ a0 = a, (a0 ' + (x0 -1l a)) -1l b = ((a0 ' + x0) -1l b) -1l a

Rule argument

pos: left 1 terms: a0, b

Additional information

Used terms: a0, b