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Proof step 28 from proof diff-sub-size (
previous step
)
#(s0 \ s) = 0, # s = # s0, a ∈ s0, a0 ∈ s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0, ∀ a1. ¬ a1 ∈ s0 \ s ⊦ 0 - 1 = # s0 - # s0 + 1, a ∈ s
User applied rule all left
And closed the goal!
Rule argument
pos: left 6 terms: a abort: F discard: F computed: F
Additional information
Used simplifier rules: Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2; Used terms: a