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Proof step 3 from proof union-size-not-less-01 (previous step)

m + n < #(s0 ∪ s1), # s0 ≤ m, # s1 ≤ n ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ #(s0 ∪ s1) ≤ # s0 + # s1
Proof step 5:
m + n < #(s0 ∪ s1), # s1 ≤ n, # s0 ≤ m, #(s0 ∪ s1) ≤ # s0 + # s1 ⊦

Rule argument

spec: <toplevel> name: union-size-le seq: ⊦ #(s0 ∪ s1) ≤ # s0 + # s1 subst: [s0, s1] → [s0, s1]

Additional information

Used terms: s0, s1