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Proof step 97 from proof n2s-s2n-identity (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 98:
⊦ char ≠ char0 → pos(char, char0 ' + chars) = pos(char, chars) + 1
Proof step 99:
'6' ≠ '0', ∀ m1. m1 < (((((0 + 1) + 1) + 1) + 1) + 1) + 1 → true → s→n(n→s(m1)) = m1 ⊦ pos('6', '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ') + 1 = 6

Rule argument

spec: list-del inst: charlist name: pos-n seq: ⊦ char ≠ char0 → pos(char, char0 ' + chars) = pos(char, chars) + 1 subst: [chars, char0, char] → [ '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ', '0', '6'] flags: rewrite

Additional information

Used terms: '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ', '0', '6'