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Proof step 5 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 cut applied rule cut formula

And obtained the following new goals

Proof step 6:
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
Proof step 9:
∀ 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

fma: a0 = a

Additional information

Used formulas: a0 = a