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Proof step 5 from proof inj-put-01 (previous step)

∀ a0. (a0 ∈ st[a, d][b, d0] ↔ a0 ∈ st[b, d0]) ∧ st[a, d][b, d0][a0] = st[b, d0][a0] ⊦ a = b, a ∈ st ∧ st[a] = d

User applied rule all left

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st);%At-other: ⊦ a ≠ b → st[b, d][a] = st[a];%At-same: ⊦ st[a, d][a] = d;%ins: ⊦ st[a] = d ∧ a ∈ st → st[a, d] = st;%notmemlookup: ⊦ ¬ a ∈ st → st[a] = (∅)[a];%In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st; Used terms: a