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Proof step 5 from proof del1-del-switch (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
a = a0, ∀ a, b. (x0 -l a) -1l b = (x0 -1l b) -l a ∨ a = b ⊦ ((a0 ' + x0) -l a) -1l b = ((a0 ' + x0) -1l b) -l a, a = b
Proof step 8:
∀ a, b. (x0 -l a) -1l b = (x0 -1l b) -l a ∨ a = b ⊦ ((a0 ' + x0) -l a) -1l b = ((a0 ' + x0) -1l b) -l a, a = b, a = a0

Rule argument

fma: a = a0

Additional information

Used formulas: a = a0