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Proof step 10 from proof del-del1-switch (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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