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Proof step 38 from proof natstring-chars (previous step)

m0 < 5, ∀ m1, char. m1 < ((((m0 + 1) + 1) + 1) + 1) + 1 → char ∈ n→s(m1) → char = '0' ∨ char = '9' ∨ char = '8' ∨ char = '7' ∨ char = '6' ∨ char = '5' ∨ char = '4' ∨ char = '3' ∨ char = '2' ∨ char = '1' ⊦ ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '0', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '9', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '8', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '7', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '6', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '5', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '4', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '3', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '2', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '1'

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 39:
⊦ chars ≠ [] ∧ n ≠ 0 → chars[n] = chars.rest[n - 1]
Proof step 40:
'5' ' + '6' ' + '7' ' + '8' ' + '9' ' ≠ [] ∧ m0 ≠ 0, m0 < 5, ∀ m1, char. m1 < ((((m0 + 1) + 1) + 1) + 1) + 1 → char ∈ n→s(m1) → char = '0' ∨ char = '9' ∨ char = '8' ∨ char = '7' ∨ char = '6' ∨ char = '5' ∨ char = '4' ∨ char = '3' ∨ char = '2' ∨ char = '1' ⊦ ('5' ' + '6' ' + '7' ' + '8' ' + '9' ').rest[m0 - 1] = '0', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ').rest[m0 - 1] = '9', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ').rest[m0 - 1] = '8', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ').rest[m0 - 1] = '7', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ').rest[m0 - 1] = '6', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ').rest[m0 - 1] = '5', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ').rest[m0 - 1] = '4', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ').rest[m0 - 1] = '3', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ').rest[m0 - 1] = '2', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ').rest[m0 - 1] = '1'

Rule argument

spec: list-del inst: charlist name: get-rec seq: ⊦ chars ≠ [] ∧ n ≠ 0 → chars[n] = chars.rest[n - 1] subst: [n, chars] → [m0, '5' ' + '6' ' + '7' ' + '8' ' + '9' '] flags: rewrite

Additional information

Used simplifier rules: ⊦ char ' + chars ≠ [];% ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1;% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ (char ' + chars)[0] = char; Used terms: m0, '5' ' + '6' ' + '7' ' + '8' ' + '9' '