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Proof step 12 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, a0 = 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 13:
(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

Rule argument

Additional information

Used simplifier rules: del1-n: ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a);%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list