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Proof step 1 from proof subseteq-rem-del (conclusion)

⊦ a ∈ s0 ∧ s0 ⊆ s1 → s1 \ (s0 -- a) = (s1 \ s0) ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ s0, s0 ⊆ s1, a ∈ s1 ⊦ s1 = ∅, # s0 = 0, s1 \ (s0 -- a) = (s1 \ s0) ++ a, s0 = ∅, # s1 = 0

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%sub-in: ⊦ a ∈ s ∧ s ⊆ s0 → a ∈ s0; (forward) from specification set-basic