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Proof step 4 from proof union-sizeeq (previous step)

#(s ∪ s0) = # s, a ∈ s0, a0 ∈ s0 ⊦ a ∈ s

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 5:
⊦ s ⊆ s0 ∧ # s = # s0 → s = s0
Proof step 6:
#(s ∪ s0) = # s, a0 ∈ s0, a ∈ s0, s ⊆ s ∪ s0 ∧ # s = #(s ∪ s0) → s = s ∪ s0 ⊦ a ∈ s

Rule argument

spec: set-basic name: subsize-eq seq: ⊦ s ⊆ s0 ∧ # s = # s0 → s = s0 subst: [s, s0] → [s, s ∪ s0]

Additional information

Used terms: s, s ∪ s0