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Proof step 13 from proof bst-branch-strong (conclusion)

⊦ bst(branch(t0, b, t1)) ↔ (∀ a. a ∈ t0 → a < b) ∧ (∀ a. a ∈ t1 → b < a) ∧ (∀ a. ¬ (a ∈ t0 ∧ a ∈ t1)) ∧ (∀ a, c. a ∈ t0 ∧ c ∈ t1 → a < c) ∧ bst(t0) ∧ bst(t1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ bst(branch(t0, b, t1)) ↔ (∀ a. a ∈ t0 → a < b) ∧ (∀ a. a ∈ t1 → b < a) ∧ (∀ a. ¬ a ∈ t0 ∨ ¬ a ∈ t1) ∧ (∀ a, c. a ∈ t0 ∧ c ∈ t1 → a < c) ∧ bst(t0) ∧ bst(t1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]