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Proof step 33 from proof diff-sub-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
a0 ∈ s, ∀ a1. a1 = a ∨ a1 ∈ s → a1 ∈ s0 ⊦ #(s0 \ s) = # s0 - #(s ++ a), # s = 0, s = ∅, a ∈ s0 \ s, a0 ∈ s0

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)