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Theorem base for Specification store+set

This is a listing of all goals defined over the current specification. Axioms are yellow, proof obligations purple and theorems green. Click on individual names to get more information on the proof.

Name: Sequent: Type:
⊦ a ∈ dom st ↔ a ∈ st An Axiom
⊦ st \ ∅ = st An Axiom
⊦ st \ (s ++ a) = (st \ s) -- a An Axiom
⊦ d ∈ st ↔ (∃ a. a ∈ st ∧ st[a] = d) An Axiom
⊦ st | s = st \ (dom st \ s) An Axiom
⊦ ¬ a ∈ st0 ∧ a ∈ st1 → (st0 ⊕ st1)[a] = st1[a] An Axiom
⊦ a ∈ st0 → (st0 ⊕ st1)[a] = st0[a] An Axiom
⊦ a ∈ st0 ⊕ st1 ↔ a ∈ st0 ∨ a ∈ st1 An Axiom
⊦ ¬ d0 ∈ st → ¬ d0 ∈ st -- a A User Theorem
⊦ ¬ d0 ∈ st ∧ d0 ≠ d1 → ¬ d0 ∈ st[b, d1] -- a A User Theorem
⊦ ¬ d0 ∈ st ∧ d0 ≠ d1 ∧ d0 ≠ d2 → ¬ d0 ∈ st[b, d1][c, d2] -- a A User Theorem
⊦ ¬ a ∈ st → (d0 ∈ st[a, d1] ↔ d0 ∈ st ∨ d0 = d1) A User Theorem
⊦ ¬ d0 ∈ st ∧ d0 ≠ d1 → ¬ d0 ∈ (st -- a)[b, d1] A User Theorem
⊦ ¬ d0 ∈ st ∧ d0 ≠ d1 ∧ d0 ≠ d2 → ¬ d0 ∈ (st[b, d1] -- a)[c, d2] A User Theorem
⊦ d0 = d1 → d0 ∈ st[a, d1] A User Theorem
⊦ d ∈ st[a, d] A User Theorem
⊦ d0 ≠ d1 ∧ ¬ d0 ∈ st → ¬ d0 ∈ st[a, d1] A User Theorem
⊦ d0 ≠ d1 ∧ d0 ≠ d2 ∧ ¬ d0 ∈ st → ¬ d0 ∈ st[a, d1][b, d2] A User Theorem
⊦ d ≠ d0 ∧ d0 ≠ d1 ∧ d0 ≠ d2 ∧ ¬ d0 ∈ st → ¬ d0 ∈ st[a, d1][b, d2][c, d] A User Theorem
⊦ (st \ s) \ s0 = (st \ s0) \ s A User Theorem
⊦ (st -- a) \ s = (st \ s) -- a A User Theorem
⊦ ∅ \ s = ∅ A User Theorem
⊦ a ∈ st \ s ↔ a ∈ st ∧ ¬ a ∈ s A User Theorem
⊦ ¬ a ∈ s → (st \ s)[a] = st[a] A User Theorem
⊦ a ∈ s → (st \ s)[a] = (∅)[a] A User Theorem
⊦ st[a, d] \ (s ++ a) = st \ (s ++ a) A User Theorem
⊦ st \ (s ++ a) = (st -- a) \ s A User Theorem
a ∈ st ⊦ st[a] ∈ st A User Theorem
⊦ ¬ a ∈ st → st \ (s ++ a) = st \ s A User Theorem
⊦ a ∈ st → (st \ (s ++ a)) -- a = st \ (s ++ a) A User Theorem
⊦ a ≠ b ∧ ¬ b ∈ s → (st \ (s ++ a))[b] = st[b] A User Theorem
⊦ a ∈ st → dom st ++ a = dom st A User Theorem
⊦ dom(st -- a) = dom st -- a A User Theorem
⊦ dom ∅ = ∅ A User Theorem
⊦ dom st[a, d] = dom st ++ a A User Theorem
⊦ dom st = ∅ ↔ st = ∅ A User Theorem
⊦ dom(st \ s) = dom st \ s A User Theorem
⊦ s ⊆ dom st → dom(st | s) = s A User Theorem
⊦ dom(st0 ⊕ st1) = dom st0 ∪ dom st1 A User Theorem
⊦ ∅ ⊕ st = st A User Theorem
⊦ a ∈ st → (a ∈ dom st ∩ s ↔ a ∈ s) A User Theorem
⊦ st ⊕ ∅ = st A User Theorem
⊦ st0[a, d] ⊕ st1 = (st0 ⊕ st1)[a, d] A User Theorem
⊦ ¬ a ∈ s → st[a, d] \ s = (st \ s)[a, d] A User Theorem
⊦ st[a, d] \ s = (a ∈ s ⊃ st \ s;(st \ s)[a, d]) A User Theorem
⊦ a ∈ s → st[a, d] \ s = st \ s A User Theorem
⊦ a ∈ st ∧ a ∈ s → (st | s)[a] = st[a] A User Theorem
⊦ st | ∅ = ∅ A User Theorem
⊦ a ∈ st | s ↔ a ∈ st ∧ a ∈ s A User Theorem
⊦ dom st ∩ s = ∅ → st \ s = st A User Theorem
⊦ dom st ∩ s = ∅ → st[a, d] \ (s -- a) = st[a, d] A User Theorem