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Proof step 7 from proof size-union (previous step)

∀ s0. (∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s0) → #(s1 ∪ s0) = # s1 + # s0, a ∈ s0 ∨ a ∈ s1, ∀ a0. a0 ≠ a ∧ ¬ a0 ∈ s1 ∨ ¬ a0 ∈ s0 ⊦ #(s0 ∪ s1) = #(s1 ++ a) + # s0

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 8:
a ∈ s1, ∀ s0. (∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s0) → #(s1 ∪ s0) = # s1 + # s0, a ∈ s0 ∨ a ∈ s1, ∀ a0. a0 ≠ a ∧ ¬ a0 ∈ s1 ∨ ¬ a0 ∈ s0 ⊦ #(s0 ∪ s1) = #(s1 ++ a) + # s0
Proof step 12:
∀ s0. (∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s0) → #(s1 ∪ s0) = # s1 + # s0, a ∈ s0 ∨ a ∈ s1, ∀ a0. a0 ≠ a ∧ ¬ a0 ∈ s1 ∨ ¬ a0 ∈ s0 ⊦ a ∈ s1, #(s0 ∪ s1) = #(s1 ++ a) + # s0

Rule argument

fma: a ∈ s1

Additional information

Used formulas: a ∈ s1