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Proof step 13 from proof sw-02 (
previous step
)
(x0 -1l a0) -1l b = (x0 -1l b) -1l a0, ∀ a, b. (x0 -1l a) -1l b = (x0 -1l b) -1l a ⊦ b = a0, (x0 -1l a) -1l b = (x0 -1l b) -1l a
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: a, b
Additional information
Used terms: a, b