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Proof step 6 from proof eq-objs (
previous step
)
((r => H[r] * P0))(H), ((r => o1 * P1))(H) ⊦ H[r] = o1
User applied rule apply lemma
And obtained the following new goals
Proof step 7
:
⊦ ((r => o0 * P))(H) → H[r] = o0
Rule argument
spec: <toplevel> name: at-rw seq: ⊦ ((r => o0 * P))(H) → H[r] = o0 subst: [r, o0, P, H] → [r, o1, P1, H]
Additional information
Used terms: r, o1, P1, H