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Proof step 11 from proof bst-branch-strong (previous step)

⊦ (∀ a. a ∈ t0 → a < b) ∧ (∀ a. a ∈ t1 → b < a) ∧ bst(t0) ∧ bst(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 10:
bst(t0), bst(t1), ∀ a. a ∈ t0 → a < b, ∀ a. a ∈ t1 → b < a ⊦ (¬ a ∈ t0 ∨ ¬ a ∈ t1) ∧ (a0 ∈ t0 ∧ c ∈ t1 → a0 < c)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]