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Proof step 4 from proof in-01 (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: b = a0

Additional information

Used formulas: b = a0