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Proof step 5 from proof subrec-01 (previous step)

a0 ∈ y, ∀ a, y. x0 ⊆m a ' + y ↔ #oc(a, x0) ≤ #oc(a, y) + 1 ∧ x0 -1l a ⊆m y ⊦ a = a0, x0 ⊆m a ' + (y -1l a0) ↔ #oc(a, x0) ≤ #oc(a, y) + 1 ∧ x0 -1l a ⊆m y -1l a0

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a, y -1l a0 abort: F discard: F computed: T

Additional information

Used simplifier rules: del1-01: ⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x); Used terms: a, y -1l a0