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Proof step 3 from proof bmin-sub-05 (previous step)

s.max ∈ s, s.min ∈ s, s.max ∈ s0, s.min ∈ s0, s0.min ∈ s0, s0.max ∈ s0, s ⊆ s0 ⊦ # s = 0, # s0 = 0, s = ∅, s0 = ∅, s.min < s0.min, s0.max < s.min, s.max < s0.min, s0.max < s.max, s -- s.min ⊆ s0 -- s0.min

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: set-basic name: Subset seq: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2) subst: [s2, s1] → [s0 -- s0.min, s -- s.min] flags: <all paths>

Additional information

Used simplifier rules: Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%min-less: ⊦ a ≠ s.min ∧ a ∈ s → s.min < a;% ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s; (ws)%sub-in: ⊦ a ∈ s ∧ s ⊆ s0 → a ∈ s0; (forward) from specification set-basic% ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss) Used terms: s0 -- s0.min, s -- s.min