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Proof step 11 from proof at-rw (previous step)

f(r) = H0[r], ((r => o))(H0), P(H), r ∈ rs, H0 ⊥ H ⊦ # rs = 0, H0[r] = o, rs = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ ((r => o))(H) ↔ H = (∅ ++ r)[r, o] ∧ r ≠ null

Rule argument

spec: maplet name: maplet seq: ⊦ ((r => o))(H) ↔ H = (∅ ++ r)[r, o] ∧ r ≠ null subst: [H, o, r] → [H0, o, r] flags: <all paths>

Additional information

Used simplifier rules: ⊦ H[r0, o][r] = (r = r0 ∧ r0 ∈ H ⊃ o;H[r]); (s)% ⊦ ¬ r ∈ ∅; (s)%r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H; (s) Used terms: H0, o, r