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Proof step 13 from proof diff-sub-size (previous step)

a1 ∈ s, a0 ∈ s0, a ∈ s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ #(s0 \ s) - 1 = # s0 - # s + 1, a ∈ s, a1 ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
a1 ∈ s, a0 ∈ s0, a ∈ s0, ∀ a2. a2 ∈ s → a2 ∈ s0 ⊦ # s = 0, #(s0 \ s) - 1 = # s0 - # s + 1, s = ∅, a ∈ s, a1 ∈ 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)