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Proof step 11 from proof union-size-03 (previous step)

# s0 + # s = n0 + #(s0 ∩ s) ∧ #(s0 ∩ s) ≤ # s0 + # s, #(s0 ∪ s) = n0 ⊦ #(s ∪ s0) + 1 = # s0 + # s + 1 - #(s0 ∩ s), a ∈ s, a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# s0 + # s = #(s0 ∪ s) + #(s0 ∩ s), #(s0 ∩ s) ≤ # s0 + # s ⊦ #(s ∪ s0) = # s0 + # s - #(s0 ∩ s), a ∈ s, a ∈ s0

Rule argument

Additional information

Used simplifier rules: eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat