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

# s0 + # s + 1 < #(s ∪ s0), a ∈ s0, a0 ∈ s0, ∀ s0. ¬ # s0 + # s < #(s0 ∪ s) ⊦ a ∈ s

User applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: s0 abort: F discard: F computed: F

Additional information

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