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Proof step 45 from proof natstring-chars (previous step)

'6' ' + '7' ' + '8' ' + '9' ' ≠ [] ∧ m ≠ 0, m < 4, ∀ m1, char. m1 < (((((m + 1) + 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' ⊦ ('6' ' + '7' ' + '8' ' + '9' ').rest[m - 1] = '0', ('6' ' + '7' ' + '8' ' + '9' ').rest[m - 1] = '9', ('6' ' + '7' ' + '8' ' + '9' ').rest[m - 1] = '8', ('6' ' + '7' ' + '8' ' + '9' ').rest[m - 1] = '7', ('6' ' + '7' ' + '8' ' + '9' ').rest[m - 1] = '6', ('6' ' + '7' ' + '8' ' + '9' ').rest[m - 1] = '5', ('6' ' + '7' ' + '8' ' + '9' ').rest[m - 1] = '4', ('6' ' + '7' ' + '8' ' + '9' ').rest[m - 1] = '3', ('6' ' + '7' ' + '8' ' + '9' ').rest[m - 1] = '2', ('6' ' + '7' ' + '8' ' + '9' ').rest[m - 1] = '1'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
m < 4, ∀ m1, char. m1 < (((((m + 1) + 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' ⊦ m = 0, ('7' ' + '8' ' + '9' ')[m - 1] = '0', ('7' ' + '8' ' + '9' ')[m - 1] = '9', ('7' ' + '8' ' + '9' ')[m - 1] = '8', ('7' ' + '8' ' + '9' ')[m - 1] = '7', ('7' ' + '8' ' + '9' ')[m - 1] = '6', ('7' ' + '8' ' + '9' ')[m - 1] = '5', ('7' ' + '8' ' + '9' ')[m - 1] = '4', ('7' ' + '8' ' + '9' ')[m - 1] = '3', ('7' ' + '8' ' + '9' ')[m - 1] = '2', ('7' ' + '8' ' + '9' ')[m - 1] = '1'

Rule argument

Additional information

Used simplifier rules: ⊦ char ' + chars ≠ [];% ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1;% ⊦ (char ' + chars).rest = chars;