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Proof step 7 from proof in-def (
previous step
)
∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z), a = a0 ∨ a ∈ x0 ⊦ a = a0, ∃ y, z. a0 ' + x0 = y + a ' + z
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 8
:
a ∈ x0, ∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z) ⊦ a = a0, ∃ y, z. a0 ' + x0 = y + a ' + z
Rule argument
Additional information