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Proof step 4 from proof in (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
b = a0, a ∈ x0, ∀ a, b. a ∈ x0 → a = b ∨ a ∈ x0 -l b ⊦ a = b, a = a0, a ∈ (a0 ' + x0) -l b
Proof step 7:
a ∈ x0, ∀ a, b. a ∈ x0 → a = b ∨ a ∈ x0 -l b ⊦ b = a0, a = b, a = a0, a ∈ (a0 ' + x0) -l b

Rule argument

fma: b = a0

Additional information

Used formulas: b = a0