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Proof step 2 from proof bst-branch-strong (previous step)

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

Heuristic pl case distinction applied rule case distinction

And closed the goal!

Rule argument

[CASEDARG : pos: right 3, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ a < a], [CSIMPFORWARD : ⊦ a < b ∧ b < c → a < c], [CSIMPFORWARD : ⊦ ¬ b < a ∧ b < c → a < c])]]