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Proof step 46 from proof natstring-chars (previous step)

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'

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 47:
m = m0 + 1, 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' ')[m0] = '0', ('7' ' + '8' ' + '9' ')[m0] = '9', ('7' ' + '8' ' + '9' ')[m0] = '8', ('7' ' + '8' ' + '9' ')[m0] = '7', ('7' ' + '8' ' + '9' ')[m0] = '6', ('7' ' + '8' ' + '9' ')[m0] = '5', ('7' ' + '8' ' + '9' ')[m0] = '4', ('7' ' + '8' ' + '9' ')[m0] = '3', ('7' ' + '8' ' + '9' ')[m0] = '2', ('7' ' + '8' ' + '9' ')[m0] = '1'

Rule argument

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

Additional information

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