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Proof step 34 from proof orderedls-all (previous step)

m < n, m < # y, n ≤ # y, ∀ m, n. m < n ∧ ¬ 1 + # y < n → (a ' + a0 ' + y)[m] < (a ' + a0 ' + y)[n] ⊦ y = [], (a0 ' + y)[m] < (a0 ' + y)[n], ordered<(a0 ' + y)

User applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: m + 1, n + 1 abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ m + n < m + n0 ↔ n < n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ (a ' + x)[n + 1] = x[n]; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: m + 1, n + 1