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Proof step 26 from proof size-exgr (
previous step
)
#(s1 ++ a) < # s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ a ∈ s0, ∃ a0. a0 ∈ s0 ∧ a0 ≠ a ∧ ¬ a0 ∈ s1
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 27
:
#(s1 ++ a) < # s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ a ∈ s0, ∃ a0. a0 ∈ s0 ∧ ¬ a0 ∈ s1
Rule argument
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