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Proof step 31 from proof natstring-chars (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m0, m] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m0, m