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

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

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 3:
⊦ x ≠ [] → (∃ y, a. x = y + a ')
Proof step 4:
z ≠ [] → (∃ y, a. z = y + a '), a ' + x + b ' = y + c ' + a0 ' + z, c ∈ z, a ∈ y ∨ a = a0 ∨ a ∈ z, b ∈ y ∨ b = a0 ∨ b ∈ z ⊦ a = b, a ∈ x, b ∈ x, dups(x)

Rule argument

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

Additional information

Used terms: z