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

#(s ∪ s0) < # s ⊦ # s = 0, s = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ s ⊆ s0 → ¬ # s0 < # s

Rule argument

spec: set-basic inst: name: sub-size seq: ⊦ s ⊆ s0 → ¬ # s0 < # s subst: [s, s0] → [s, s ∪ s0] flags: rewrite

Additional information

Used simplifier rules: unsub: ⊦ s ⊆ s ∪ s0;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s; Used terms: s, s ∪ s0