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Proof step 6 from proof leftmost-left (previous step)

⊦ ((t0 = leaf ⊃ a; t0 .leftmost)) = a, ((t0 = leaf ⊃ a; t0 .leftmost)) ∈ t0

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 5:
∀ a. ((t = leaf ⊃ a; t .leftmost)) = a ∨ ((t = leaf ⊃ a; t .leftmost)) ∈ t, ∀ a. ((t1 = leaf ⊃ a; t1 .leftmost)) = a ∨ ((t1 = leaf ⊃ a; t1 .leftmost)) ∈ t1 ⊦ ((branch(t, a0, t1) = leaf ⊃ a; branch(t, a0, t1) .leftmost)) = a, ((branch(t, a0, t1) = leaf ⊃ a; branch(t, a0, t1) .leftmost)) ∈ branch(t, a0, t1)

Rule argument

[term-arg: t0]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : tree generated by leaf, branch;])]]