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Proof step 36 from proof orderedls-all (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: 0, 1 abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ (a ' + x)[0] = a; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1]; (s) Used terms: 0, 1