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Proof step 10 from proof not-always-decrease (previous step)

∀ a1. a1 ≪ a0 → \A ∀ A, a. a1 ≠ A ∨ ¬ □ ⟨a := A⟩ (◇ A ≪ a) ∨ a ≠ A, □ ⟨a0 := A⟩ (◇ A ≪ a0), ◇ (a = A ∧ A ≪ a0), a0 = A ⊦

User applied rule all left

And obtained the following new goals

Proof step 11:
a ≪ a0 → \A ∀ A, a0. a ≠ A ∨ ¬ □ ⟨a0 := A⟩ (◇ A ≪ a0) ∨ a0 ≠ A, □ ⟨a0 := A⟩ (◇ A ≪ a0), ◇ (a = A ∧ A ≪ a0), a0 = A ⊦

Rule argument

pos: left 1 terms: a abort: F discard: T computed: F

Additional information

Used terms: a