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Proof step 17 from proof diff-sub-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
#(s0 \ s) = # s0 - # s, # s0 < # s, a0 ∈ s0, a ∈ s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ # s0 - # s - 1 = # s0 - # s + 1, a ∈ s

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat