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Proof step 7 from proof leftmost-left (previous step)

t = branch(t0, a, t1) ⊦ t = leaf, t .leftmost = a, t .leftmost ∈ t0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ branch(t0, a, t1) .leftmost = ((t0 = leaf ⊃ a; t0 .leftmost))
Proof step 6:
⊦ ((t0 = leaf ⊃ a; t0 .leftmost)) = a, ((t0 = leaf ⊃ a; t0 .leftmost)) ∈ t0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ leaf ≠ branch(t, a, t0)])]]