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Proof step 5 from proof Data-in-other (previous step)

a0 ∈ st ⊦ d1 = st[a0], a ∈ st, ∃ a1. a1 ∈ st[a, d1] ∧ st[a, d1][a1] = st[a0]

User applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: a0 abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ b ∈ st ∧ ¬ a ∈ st → st[a, d][b] = st[b]; (s)% ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st; (ws) Used terms: a0