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Proof step 2 from proof lasttwo (previous step)

a0 ∈ x, a ∈ x ⊦ ∃ b, y, z. x = y + b ' + z ∧ (b = a ∨ b = a0) ∧ ¬ a ∈ z ∧ ¬ a0 ∈ z

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ a ∈ x → (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ z)
Proof step 4:
a ∈ x, a0 ∈ x, a0 ∈ x → (∃ y, z. x = y + a0 ' + z ∧ ¬ a0 ∈ z) ⊦ ∃ b, y, z. x = y + b ' + z ∧ (b = a ∨ b = a0) ∧ ¬ a ∈ z ∧ ¬ a0 ∈ z

Rule argument

spec: <toplevel> name: lastoc seq: ⊦ a ∈ x → (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ z) subst: [a, x] → [a0, x]

Additional information

Used terms: a0, x