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Proof step 15 from proof size-exgr (previous step)

a ∈ s0, # s1 + 1 < # s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ # s0 = 0, a ∈ s1, ∃ a0. a0 ∈ s0 ∧ a0 ≠ a ∧ ¬ a0 ∈ s1

User applied rule constructor cut

And obtained the following new goals

Proof step 16:
s1 = ∅, a ∈ s0, # s1 + 1 < # s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ # s0 = 0, a ∈ s1, ∃ a0. a0 ∈ s0 ∧ a0 ≠ a ∧ ¬ a0 ∈ s1
Proof step 21:
s1 = s ++ a1 ∧ s1 ≠ s, a ∈ s0, # s1 + 1 < # s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ # s0 = 0, a ∈ s1, ∃ a0. a0 ∈ s0 ∧ a0 ≠ a ∧ ¬ a0 ∈ s1

Rule argument

var: s1

Additional information

Used simplifier rules: gen: ⊦ ;