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Proof step 5 from proof leftmost-left (previous step)

∀ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ branch(t0, a, t1) .leftmost = ((t0 = leaf ⊃ a; t0 .leftmost))
Proof step 3:
⊦ a ∈ branch(t0, b, t1) ↔ ¬ ¬ (a = b ∨ a ∈ t0 ∨ a ∈ t1)
Proof step 4:
∀ 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 ⊦ ((t = leaf ⊃ a0; t .leftmost)) = a, ((t = leaf ⊃ a0; t .leftmost)) = a0, ((t = leaf ⊃ a0; t .leftmost)) ∈ t1, ((t = leaf ⊃ a0; t .leftmost)) ∈ t

Rule argument

[emptyarg]

Additional information

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