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Proof step 112 from proof n2s-s2n-identity (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 113:
⊦ char ≠ char0 → pos(char, char0 ' + chars) = pos(char, chars) + 1

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] → ['6' ' + '7' ' + '8' ' + '9' ', '5', '6'] flags: rewrite

Additional information

Used simplifier rules: pos-y: ⊦ pos(char, char ' + chars) = 0; from specification list-del[charlist]%a: ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1; from specification list[charlist]%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: '6' ' + '7' ' + '8' ' + '9' ', '5', '6'