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Proof step 1 from proof lasttwo (conclusion)
a ∈ x, a0 ∈ x ⊦ ∃ b, y, z. x = y + b ' + z ∧ (b = a ∨ b = a0) ∧ ¬ a ∈ z ∧ ¬ a0 ∈ z
System applied rule normalize
And obtained the following new goals
Proof step 2
:
a0 ∈ x, a ∈ x ⊦ ∃ b, y, z. x = y + b ' + z ∧ (b = a ∨ b = a0) ∧ ¬ a ∈ z ∧ ¬ a0 ∈ z
Rule argument
Additional information