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Proof step 11 from proof not-always-decrease (previous step)

a ≪ a0 → \A ∀ A, a0. a ≠ A ∨ ¬ □ ⟨a0 := A⟩ (◇ A ≪ a0) ∨ a0 ≠ A, □ ⟨a0 := A⟩ (◇ A ≪ a0), ◇ (a = A ∧ A ≪ a0), a0 = A ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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