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Proof step 15 from proof n2s-s2n-identity (previous step)

m < 10, ∀ m1. m1 < m → true → s→n(n→s(m1)) = m1 ⊦ s→n(('0' + '1' + '2' + '3' + '4' + '5' + '6' + '7' + ('8' + '9'))[m] + "") = m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
m < 10, ∀ m1. m1 < m → true → s→n(n→s(m1)) = m1 ⊦ s→n(('0' ' + '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m] ') = m

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%assoc: ⊦ (str1 + str2) + str3 = str1 + str2 + str3; from specification string-append%empty-01: ⊦ str1 + "" = str1; from specification string-append%chartostring-01: ⊦ char + str = char ' + str; from specification string-append%c: ⊦ char + chars = char ' + chars; from specification list[charlist]%Two: ⊦ char + char0 = char ' + char0 '; from specification list[charlist]