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Proof step 25 from proof not-always-decrease (previous step)

N = N'' +1 until A ≪ a, m = N, \A ∀ N, A, a. N < m → N = N'' +1 until A ≪ a → (∃ a1. ◇ (a1 = A ∧ A ≪ a)) ⊦ ∃ a1. ◇ (a1 = A ∧ A ≪ a)

User applied rule step

And obtained the following new goals

Proof step 26:
a0 ≪ a ∨ n = N +1 ∧ N = N'' +1 until A ≪ a, m = n, boolvar → a0 = a2 ∧ n = n0, \A ∀ N, A, a. N < m → N = N'' +1 until A ≪ a → (∃ a1. ◇ (a1 = A ∧ A ≪ a)) ⊦ ∃ a1. a1 = a0 ∧ a0 ≪ a ∨ ◇ (a1 = A ∧ A ≪ a)

Rule argument

args:[positions: , ]

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