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Proof step 6 from proof lasttwo (previous step)

a ∈ z ⊦ a0 ∈ z, ∃ b, y0, z0. y + a0 ' + z = y0 + b ' + z0 ∧ (b = a ∨ b = a0) ∧ ¬ a ∈ z0 ∧ ¬ a0 ∈ z0

User applied rule insert lemma

And obtained the following new goals

Proof step 7:
⊦ a ∈ x → (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ z)
Proof step 8:
a ∈ z → (∃ y, z0. z = y + a ' + z0 ∧ ¬ a ∈ z0), a ∈ z ⊦ a0 ∈ z, ∃ b, y0, z0. y + a0 ' + z = y0 + b ' + z0 ∧ (b = a ∨ b = a0) ∧ ¬ a ∈ z0 ∧ ¬ a0 ∈ z0

Rule argument

spec: list inst: name: lastoc seq: ⊦ a ∈ x → (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ z) subst: [a, x] → [a, z] flags: current, []

Additional information

Used terms: a, z