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Proof step 53 from proof natstring-chars (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: ⊦ (char ' + chars)[0] = char;% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ (char ' + chars).rest = chars;% ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1;% ⊦ char ' + chars ≠ [];% ⊦ m ≠ 0 → (m < 2 ↔ m = 1);% ⊦ char '[0] = char; Used terms: m, '8' ' + '9' '