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Proof step 2 from proof subsize-eq-01 (previous step)

# s = # s0, s ⊆ s0 ⊦ s = s0

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ s ⊆ s0 ∧ # s = # s0 → s = s0

Rule argument

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

Additional information

Used terms: s, s0