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Proof step 3 from proof inapp-01 (previous step)

a ' + x + b ' = y + c ' + a0 ' + z, c ∈ z, c = a ∨ c ∈ x ∨ c = b, a0 = a ∨ a0 ∈ x ∨ a0 = b, a ∈ y ∨ a = a0 ∨ a ∈ z, b ∈ y ∨ b = a0 ∨ b ∈ z ⊦ a = b, dups(x), b ∈ x, a ∈ x

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 4:
⊦ x ≠ [] → (∃ y, a. x = y + a ')
Proof step 5:
a ' + x + b ' = y + c ' + a0 ' + z, c ∈ z, z ≠ [] → (∃ y, a. z = y + a '), b ∈ y ∨ b = a0 ∨ b ∈ z, a ∈ y ∨ a = a0 ∨ a ∈ z, a0 = a ∨ a0 ∈ x ∨ a0 = b, c = a ∨ c ∈ x ∨ c = b ⊦ a = b, dups(x), a ∈ x, b ∈ x

Rule argument

spec: list name: lastdiv seq: ⊦ x ≠ [] → (∃ y, a. x = y + a ') subst: [x] → [z]

Additional information

Used terms: z