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Proof step 3 from proof charn1 (previous step)

# "1" = 1 ⊦ pos("1".char1, '0' + '1' + '2' + '3' + '4' + '5' + '6' + '7' + ('8' + '9')) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ char + chars = char ' + chars;% ⊦ char + char0 = char ' + char0 ';% ⊦ # "" = 0;%str ≠ "" ⊦ # (* str) = # str.rstring + 1;