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Proof step 10 from proof ins-max (previous step)

a0 < a, ordered≤(a0 ' + x), ∀ a1. ins≤(a1, x) = x + a1 ' ∨ a1 < max(x) ⊦ ins≤(a, x) = x + a ', a < max(a0 ' + x), ordered≤(a ' + a0 ' + x)

User applied rule all left

And obtained the following new goals

Proof step 11:
a0 < a, ordered≤(a0 ' + x), ins≤(a, x) = x + a ' ∨ a < max(x), ∀ a1. ins≤(a1, x) = x + a1 ' ∨ a1 < max(x) ⊦ ins≤(a, x) = x + a ', a < max(a0 ' + x), ordered≤(a ' + a0 ' + x)

Rule argument

pos: left 3 terms: a abort: F discard: F computed: T

Additional information

Used terms: a