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Proof step 2 from proof orderedls-all (previous step)

ordered<(x0) ↔ (∀ m, n. m < n ∧ n < # x0 → x0[m] < x0[n]) ⊦ ordered<(a + x0) ↔ (∀ m, n. m < n ∧ n < #(a + x0) → (a + x0)[m] < (a + x0)[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
ordered<(x0) ↔ (∀ m, n. m < n ∧ n < # x0 → x0[m] < x0[n]) ⊦ ordered<(a ' + x0) ↔ (∀ m, n. m < n ∧ ¬ # x0 < n → (a ' + x0)[m] < (a ' + x0)[n])

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%len-02: ⊦ # a ' = 0 +1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%c: ⊦ a + x = a ' + x; from specification list