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Proof step 10 from proof bst-branch-strong (previous step)

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

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 9:
bst(t0), bst(t1), ∀ a. a ∈ t0 → a < b, c ∈ t1 → b < c, ∀ a. a ∈ t1 → b < a ⊦ (¬ a ∈ t0 ∨ ¬ a ∈ t1) ∧ (a0 ∈ t0 ∧ c ∈ t1 → a0 < c)

Rule argument

[EXRARG : pos: left 4, [QUANTTERMLIST : (c)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (c)]]