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Proof step 10 from proof eqpos (previous step)

pos(char1, str0) = pos(char2, str0), pos(char2, str0) < # str0, ∀ char0, char3. pos(char0, str0) = pos(char3, str0) ∧ pos(char3, str0) < # str0 → char0 = char3 ⊦ char2 = char, char1 = char2, char1 = char, str0 = ""

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: char1, char2

Additional information

Used terms: char1, char2