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Proof step 9 from proof del-del1-switch (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ a, b. (x0 -1l a) -l b = (x0 -l b) -1l a ∨ a = b ⊦ b = a0, a = a0, a = b, (x0 -1l a) -l b = (x0 -l b) -1l a

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);%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list