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Proof step 12 from proof gstore-dom-eq-hole (previous step)

gstore0 \ (∅ ++ n) = gstore1 \ (∅ ++ n), n ∈ gstore0, ∀ n1. n1 ∈ gstore0 ↔ n1 ∈ gstore1 ⊦ (n0 ∈ gstore1 ↔ n0 ∈ gstore0) ∧ (n0 ∈ gstore1 → gstore1[n0] = gstore0[n, gstore1[n]][n0])

User applied rule all left

And obtained the following new goals

Proof step 13:
n0 ∈ gstore0 ↔ n0 ∈ gstore1, gstore0 \ (∅ ++ n) = gstore1 \ (∅ ++ n), n ∈ gstore0 ⊦ (n0 ∈ gstore1 ↔ n0 ∈ gstore0) ∧ (n0 ∈ gstore1 → gstore1[n0] = gstore0[n, gstore1[n]][n0])

Rule argument

pos: left 3 terms: n0 abort: F discard: T computed: T

Additional information

Used terms: n0