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Proof step 12 from proof union-size-le (previous step)

a ∈ s0, # s0 + # s + 1 < #(s ∪ s0), ∀ s0. # s0 + # s < #(s0 ∪ s) → #(s0 ∪ s) = 0 ∨ s0 ∪ s = ∅ ⊦ s0 = ∅, s ∪ s0 = ∅, #(s ∪ s0) = 0, # s0 = 0, a ∈ s

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: s0 abort: F discard: F computed: F

Additional information

Used simplifier rules: unc: ⊦ s ∪ s0 = s0 ∪ s;%nls-fst: ⊦ ¬ m < n → ¬ m + m0 < n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: s0