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Proof step 2 from proof union-size-ls-in (previous step)

# s0 + # s1 ≤ #(s0 ∪ s1), a ∈ s1, a ∈ s0 ⊦ s1 = ∅, s0 = ∅, # s1 = 0, # s0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: set-union inst: name: union-size-03 seq: ⊦ #(s0 ∪ s1) = # s0 + # s1 - #(s0 ∩ s1) subst: [s1, s0] → [s1, s0] flags: rewrite, current

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;%int-empty-02: ⊦ s ∩ s0 = ∅ ∧ a ∈ s → (¬ ¬ a ∈ s0 ↔ false);%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%ls-02: n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; from specification nat%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%le-add: ⊦ m + m0 ≤ m ↔ m0 = 0; from specification nat%size-zero-02: ⊦ # s = 0 → s = ∅; (forward) from specification set-basic Used terms: s1, s0