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Proof step 11 from proof exnotempty (previous step)

a ∈ st, ∀ a0, d, st0. st ≠ st0[a0, d] ∨ a0 ∈ st0 ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a, st[a], st -- a abort: F discard: T computed: F

Additional information

Used simplifier rules: del-put: ⊦ (st -- a)[a, d] = st[a, d];%ins-same-01: ⊦ a ∈ st → st[a, st[a]] = st;%Del-notin: ⊦ ¬ a ∈ st -- a; Used terms: a, st[a], st -- a