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Proof step 57 from proof n2s-s2n-identity (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 58:
n ≠ 0 ⊦ (char ' + chars)[(* n)] = chars[n -1]
Proof step 59:
¬ m ≠ 0, m < 6, ∀ m1. m1 < (((m + 1) + 1) + 1) + 1 → true → s→n(n→s(m1)) = m1 ⊦ pos(('4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m], '0' ' + '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ') = 4 + m
Proof step 71:
m ≠ 0, m < 6, ∀ m1. m1 < (((m + 1) + 1) + 1) + 1 → true → s→n(n→s(m1)) = m1 ⊦ pos(('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m -1], '0' ' + '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ') = 4 + m

Rule argument

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

Additional information

Used terms: m, '5' ' + '6' ' + '7' ' + '8' ' + '9' ', '4'