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Proof step 25 from proof ismin (previous step)

bst(t) ∧ b ∈ t ∧ b < t .leftmost, ∀ t0, b. # t0 < # t → (bst(t0) ∧ b ∈ t0 ∧ b < t0 .leftmost → b = t0 .leftmost ∨ t0 = leaf) ⊦ b = t .leftmost ∨ t = leaf

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
b < t .leftmost, b ∈ t, bst(t), ∀ t0, b. # t0 < # t → (bst(t0) ∧ b ∈ t0 ∧ b < t0 .leftmost → b = t0 .leftmost ∨ t0 = leaf) ⊦ t = leaf, b = t .leftmost

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]