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Proof step 2 from proof isin-01 (previous step)

c ' + x + b ' = y + a ' + a ' + z, c ∈ y ∨ c = a ∨ c ∈ z ⊦ a = b, a ∈ x

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 3:
c ' + x + b ' = y + a ' + a ' + z, c ∈ y ∨ c = a ∨ c ∈ z ⊦ y = [], a = b, a ∈ x

Rule argument

fma: y = []

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;%in-empty: ⊦ ¬ a ∈ [];%inapp-10: ⊦ a ' + x = y + b ' ∧ a ≠ b → a ∈ y ∧ b ∈ x;%in-14: ⊦ a ' + x = y → a ∈ y; Used formulas: y = []