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Proof step 30 from proof diff-sub-size (previous step)

∀ s0. (∀ a. a ∈ s → a ∈ s0) → #(s0 \ s) = # s0 - # s, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ #(s0 \ s) = # s0 - #(s ++ a), a ∈ s0 \ s

User applied rule all left

And obtained the following new goals

Proof step 31:
(∀ a. a ∈ s → a ∈ s0) → #(s0 \ s) = # s0 - # s, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ #(s0 \ s) = # s0 - #(s ++ a), a ∈ s0 \ s

Rule argument

pos: left 1 terms: s0 abort: F discard: T computed: T

Additional information

Used terms: s0