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Proof step 16 from proof diff-sub-size (previous step)

#(s0 \ s) = # s0 - # s, a ∈ s0, a0 ∈ s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ # s0 - # s - 1 = # s0 - # s + 1, a ∈ s

User applied rule insert elim lemma

And obtained the following new goals

Proof step 17:
#(s0 \ s) = # s0 - # s, a ∈ s0, a0 ∈ s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ # s ≤ # s0, # s0 - # s - 1 = # s0 - # s + 1, a ∈ s
Proof step 24:
# s0 = n0 + # s ∧ # s ≤ # s0, #(s0 \ s) = n0, a ∈ s0, a0 ∈ s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ n0 - 1 = # s0 - # s + 1, a ∈ s

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [# s0, # s, n] flags:

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # s0, # s, n