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Proof step 11 from proof subset-size (previous step)

a ∈ st0, st1 -- a ⊆ st0 -- a, a ∈ st1, # st0 < # st1, ∀ st0. # st0 < # st1 → ¬ st1 ⊆ st0 ⊦ st1[a] = st0[a], st1 = ∅, st1[a, st0[a]] = ∅

User applied rule all left

And obtained the following new goals

Proof step 12:
a ∈ st0, st1 -- a ⊆ st0 -- a, a ∈ st1, # st0 < # st1, # st0[a, st1[a]] < # st1 → ¬ st1 ⊆ st0[a, st1[a]], ∀ st0. # st0 < # st1 → ¬ st1 ⊆ st0 ⊦ st1[a] = st0[a], st1 = ∅, st1[a, st0[a]] = ∅

Rule argument

pos: left 5 terms: st0[a, st1[a]] abort: F discard: F computed: F

Additional information

Used terms: st0[a, st1[a]]