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Proof step 6 from proof sw (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ a, b. (x0 -l a) -l b = (x0 -l b) -l a ⊦ (x0 -l a0) -l b = (x0 -l b) -l a0

Rule argument

Additional information

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