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Proof step 24 from proof ismin (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 1:
⊦ t ≠ leaf → t .leftmost ∈ t .left ∨ t .leftmost = t .val
Proof step 23:
t ≠ leaf → t .leftmost ∈ t .left ∨ t .leftmost = t .val, 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

[x0lemmaarg: spec-name: instance-name: lemma-name: leftmost-left antecedent: succedent: t ≠ leaf → t .leftmost ∈ t .left ∨ t .leftmost = t .val substlist: [t] → [t] add preconditions?: F is rewrite lemma?: F current spec?: T rotate lemma?: F replace all?: F paths: () dummy: () ]

Additional information

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