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Proof step 10 from proof union-size-03 (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
# s0 + # s = n0 + #(s0 ∩ s) ∧ #(s0 ∩ s) ≤ # s0 + # s, #(s0 ∪ s) = n0 ⊦ #(s ∪ s0) + 1 = # s0 + # s + 1 - #(s0 ∩ s), a ∈ s, a ∈ s0

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [# s0 + # s, #(s0 ∩ s), n] flags:

Additional information

Used simplifier rules: intersection-size: ⊦ ¬ # s0 + # s < #(s0 ∩ s);%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # s0 + # s, #(s0 ∩ s), n