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Proof step 4 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 ⊦ ((t = leaf ⊃ a0; t .leftmost)) = a, ((t = leaf ⊃ a0; t .leftmost)) = a0, ((t = leaf ⊃ a0; t .leftmost)) ∈ t1, ((t = leaf ⊃ a0; t .leftmost)) ∈ t

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 1, [QUANTTERMLIST : (a0)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (a0)]]