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Proof step 2 from proof inapp (previous step)

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

Rule argument

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

Additional information

Used terms: z